Class 11- Chapter 3 : Motion In A Straight Line

 

 MOTION IN A STRAIGHT LINE

MOTION

• Motion is the change in position of an object with time with respect to its surroundings.
• The branch of Physics which deals with the motion of objects is called  Mechanics.

Classification of Mechanics :

• Mechanics is classified into three categories :

• 1. Statics 2. Kinematics 3. Dynamics

• Statics deals with object at rest under the action of forces.

• Kinematics deals with objects in  motion without going into the causes of motion.

• Dynamics deals with objects in motion by considering the causes of motion.

POINT OBJECT

• If the size of the object is much smaller than the distance it moves, it is considered as point object.

Examples

1. A railway carriage moving without jerks between two stations.
2. A monkey sitting on top of a man cycling smoothly on a circular track.

EXTENDED OBJECT

• If the size of the object is much larger in comparison to  the distance it moves, it is considered as an extended  object.

Examples

1. An  iceberg breaking down under its own weight.
2. A Space station revolving round the Earth with a space Lab inside where a dozen Scientists are doing Research.

 

FRAME OF REFERENCE

• A frame of reference is  a place from which motion is observed and measured.
• Example: The Cartesian coordinate system with a clock – the reference point at the origin.

TYPES OF MOTION

• Based on the number of coordinates required to describe motion, motion can be classified as:

1. One dimensional motion    ( Motion in a Straight line )
2. Two-dimensional motion    ( Motion in a Plane )
3. Three-dimensional motion. ( Motion in Space )

One dimensional motion

• Motion along a straight line is called one dimensional motion or rectilinear motion.
• Only one coordinate is required to describe this motion.
• In one-dimensional motion, there are only two directions (backward and forward, upward and downward, Towards left or towards right) in which an object can move

Example:

1. A car moving on a straight road.
2. A freely falling body under gravity.

Two-dimensional motion

• Motion in a plane is called two dimensional motions.
• Two coordinates are required to represent this motion.

Example :

1. A car moving on a plane ground
2. A boat moving on a still lake

Three-dimensional motion

Motion in a space is called three dimensional motion.

Three coordinates are required to represent this motion.

Example

1. Movement of gas molecules
2. A flying bird

Position, Distance, and Displacement

In order to study how something moves, we must know where it is. This location is an object's position. To visualize position for objects moving in a straight line, you can imagine the object is on a number line. It may be placed at any point on the number line in the positive numbers or the negative numbers.  It is common to choose the original position of the object to be on the zero mark.  In making the zero mark the reference point, you have chosen a frame of reference.  The exact position of an object is the separation between the object and the reference point.

When an object moves, we often refer to the amount it moves as the distance. Distance does not need a reference point and does not need a direction. If an automobile moves 50 kilometers, the distance traveled is 50 kilometers regardless of the starting point or the direction of movement. If we wish to find the final position of the automobile, however, just having the distance traveled will not allow us to determine the final position. We need to know the starting point and the direction of the motion. The change in the position of the object is called its displacement. The displacement must include a direction because the final position may be either in the positive or negative direction along the number line from the initial position.  The displacement is a vector quantity and vectors are discussed in the chapter "Vectors".

Summary

• The length traveled by an object moving in any direction or even changing direction is called distance.
• The location of an object in a frame of reference is called position.
• For straight line motion, positions can be shown using a number line.
• The separation between original and final position is called displacement.

Questions as your Home Work

1. What is position?
2. Can two objects be the same distance from a single point but be in different positions? Why or why not?
3. What is the difference between distance and displacement?
4. Does distance have direction? Does displacement have direction?

PATH LENGTH (DISTANCE)

• The length of the path covered by an object is called path length.
• It is the total distance travelled by the object.
• Path length is a scalar quantity — a quantity that has a magnitude only and no direction.

• For example, the path length of the car moving from O to P and then from P to Q is 360+120 = 480 m.

DISPLACEMENT

• It is the change of position in a definite direction.
• Displacement is a vector quantity –have both magnitude and direction.
• It can be positive, negative or zero.
• In one dimensional motion direction, the two directions can be represented using positive (+) and negative (-) signs.
• If x1 and x2 are the positions of an object in time t1 and t2, the displacement in time interval Δt = t2 − t1 , is given by                            

                                                   Δx = x2 − x1

• If x2 > x1 , displacement is positive
• if x2 < x1 , displacement is negative.
• The magnitude of displacement may or may not be equal to the path length traversed by an object.
• If the motion of an object is along a straight line and in the same direction, the magnitude of displacement is equal to the total path length.

Uniform motion

• If an object moving along the straight line covers equal distances in equal intervals of time, it is said to be in uniform motion along a straight line.

• AVERAGE VELOCITY

  • Ratio of total displacement to the total time.
  • 
  • 
  • The SI unit for velocity is m/s or m s⁻¹
  • The unit km h–1 is used in many everyday applications
  • 
  • Average velocity is a vector quantity
  • Average velocity can be positive or negative or zero.
  • Slope of the Displacement-Time graph gives the average velocity.

  AVERAGE SPEED

  • Ratio of total path length travelled to the total time interval
  • 
  • Average speed over a finite interval of time is greater or equal to the magnitude of the average velocity
  • If the motion of an object is along a straight line and in the same direction, the magnitude of average velocity is equal to average speed.
  • SI unit of average speed is same as that of velocity.

  PROBLEM : 1

  • A car is moving along a straight line, It moves from O to P in 18 s and returns from P to Q in 6.0 s. What are the average velocity and average speed of the car in going (a) from O to P ? and (b) from O to P and back to Q ?

  

  Solution     (a).Average velocity

  

• (b) Average velocity

• 
  

• 
  

• 

  Instantaneous Velocity

  The instantaneous velocity of an object is the velocity of the object at a given moment. If the object is moving with constant velocity, then the instantaneous velocity at every moment, the average velocity, and the constant velocity are all the same.

  Position vs Time Graphs

  Consider a position versus time graph for an object starting at  t=0
   and $\begin{array}{r}x=0\end{array}$ that has a constant velocity of 80. m/s.

  

  The velocity of an object can be found from a position vs time graph. On a position vs time graph, the displacement is the vertical separation between two points and the time interval is the horizontal separation. The ratio of displacement to time interval is the average velocity. This ratio is also the slope of the line. Therefore, the slope of the straight line is the average velocity. For the motion pictured above, Slope Can be calculated as a ratio of rise and run which is 

  

  For accelerated motion (the velocity is constantly changing), the position vs time graph will be a curved line. The slope of the curved line at any point is the instantaneous velocity at that time. If we were using calculus, the slope of a curved line could be calculated. Without calculus, we approximate the instantaneous velocity at a particular point by laying a straight edge along the curved line and estimating the slope.

  

  In the image above, the red line is the position vs time graph and the blue line is an approximated slope for the line at t=2.5 second. The rise for this slope is approximately 170 m and the time interval (run) is 4.0 seconds.  Therefore, the approximated slope is 43 m/s. 

• INSTANTANEOUS VELOCITY  

• It is the velocity of the object, calculated in the shortest instant of time possible (calculated as the time interval Δt tends to zero).

where dS/dt is the derivative of displacement vector ‘S’, with respect to ‘t’. 

• The instantaneous velocity at a particular moment is calculated by substituting the corresponding time variable’s value, in the first time derivative of the displacement equation.

 

• It is the velocity of an object at an instant.
• It is the average velocity as the time interval tends to zero
• For uniform motion, velocity is the same as the average velocity at all instants.

Problem 2:

A bullet fired in space is traveling in a straight line and its equation of motion is S(t) = 4t + 6t2. If it travels for 15 seconds before impact, find the instantaneous velocity at the 11th second.

Problem 3:

A body is released to fall under the influence of gravity. Its approximate equation of motion is given by S(t) = 4.9 t2. What would be the instantaneous velocity of the body at the fifth second after release?

INSTANTANEOUS SPEED (SPEED)

• Instantaneous speed or simply speed is the magnitude of velocity
• Instantaneous speed at an instant is equal to the magnitude of the instantaneous velocity at that instant because for an infinitesimally small time interval, the motion of a particle can be approximated to be uniform.

AVERAGE ACCELERATION

• Ratio of change in velocity to time interval

• Where v2 and v1 are the instantaneous velocities or simply velocities at time t2 and t1 .
• SI unit is m/s2.
• Slope of the velocity-time graph gives average acceleration.

INSTANTANEOUS ACCELERATION (ACCELERATION)

• It is the acceleration at an instant.
• It is the average acceleration as the as the time interval tends to zero

• The instantaneous acceleration is the slope of the tangent to the v–t curve at that instant.
• Acceleration can be positive, negative or zero.
• It is a vector quantity.

Uniform Acceleration

Acceleration that does not change in time is called uniform or constant acceleration.  The velocity at the beginning of the time interval is called initial velocity, u and the velocity at the end of the time interval is called final velocity, v .  In a velocity versus time graph for uniform acceleration, the slope of the line is the acceleration. The equation that describes the curve is v = u + a t

Example 1

If an automobile with a velocity of 4.0 m/s accelerates at a rate of 4.0 m/s² for 2.5 s, what is the final velocity?

v=u+ at = 4.0 m/s+(4.0 m/s²)(2.5 s) = 4.0 m/s+10. m/s =14 m/s

Example 2

If a cart slows from 22.0 m/s with an acceleration of -2.0 m/s², how long does it require to get to 4 m/s?



t = (v−u)/a = (−18 m/s) / (−2.0 m/s²) = 9.0 s

 Suppose we plot the velocity versus time graph for an object undergoing uniform acceleration.  In this case, let us assume the object started from rest.

If the object has a uniform acceleration of 6.0 m/s² and started from rest, then each succeeding second, the velocity will increase by 6.0 m/s.  Here is the table of values and the graph.

In displacement versus time graphs, the slope of the line is the velocity of the object. In this case of a velocity versus time graph, the slope of the line is the acceleration. If you take any segment of this line and determine the Δy
 to Δx ratio, you will get 6.0 m/s² which we know to be the constant acceleration of this object.

We know from geometry that the area of a triangle is calculated by multiplying one-half the base times the height.  The area under the curve in the image above is the area of the triangle shown below. The area of this triangle would be calculated by area=(1/2)(6.0 s)(36 m/s)=108 m
.

By going back to equation 2, we know that  displacement 

S =1/2 at².

Using this equation, we can determine that  the displacement of this object in the first 6 seconds of travel is given by  

S = (1/2)(6.0 m/s²)(6.0 s)²=108 m
.

It is not coincidental that this number is the same as the area of the triangle. In fact, the area underneath the curve in a velocity versus time graph is always equal to the displacement that occurs during that time interval.

Home Work:

1. An airplane accelerates with a constant rate of 3.0 m/s² starting at a velocity of 21 m/s.  If the distance traveled during this acceleration was 535 m, what is the final velocity?
2. An car is brought to rest in a distance of 484 m using a constant acceleration of -8.0 m/s².  What was the velocity of the car when the acceleration first began?
3. An airplane starts from rest and accelerates at a constant 3.00 m/s² for 20.0 s.  What is its displacement in this time?
4. A driver brings a car to a full stop in 2.0 s. 
   1. If the car was initially traveling at 22 m/s, what was the acceleration?
   2. How far did the car travel during braking?

GRAPHS RELATED TO MOTION

POSITION-TIME GRAPH ( x –t Graph)

• It is the graph drawn taking time along x-axis and position along y-axis
• Slope of the x-t graph gives the average velocity.
• Slope of the tangent at a point in the x-t graph gives the velocity at that point.

Uses of Position –Time Graph

• To find the position at any instant
• To find the velocity at any instant
• To obtain the nature of motion

Position- time graph of stationary object

(Position- time graph of an object in uniform motion)

• 

• Position-time graph of a car

  • The car starts from rest at time t = 0 s from the origin O and picks up speed till t = 10 s and thereafter moves with uniform speed till t = 18 s. Then the brakes are applied and the car stops at  't' = 20 s and 'x' = 296 m.

  

  Position-time graph of an object moving with positive velocity

• 

  Position-time graph of an object moving with negative velocity

  

   

  Position-time graph for motion with positive acceleration

  

  Position-time graph for motion with negative acceleration

  

  Position-time graph for motion with zero acceleration

  

  PROBLEM 4

  • Calculate the average velocity between 5s and 7s from the graph.

  

  Solution

  
  
  

  
  

  PROBLEM-5

  A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.

  Solution

  

Hence, the time taken by the drunkard to fall into the pits =37s

• VELOCITY – TIME GRAPH ( v-t GRAPH)

  • A graph with velocity along Y –axis and time along X-axis.
  • The acceleration at an instant is the slope of the tangent to the v–t curve at that instant.
  • Area under the v-t graph gives the displacement.

  Uses of v-t graph

  • To find the displacement
  • To find the velocity at any time
  • To find the acceleration at any time
  • To know the nature of motion
  
  

  v-t graph of motion in positive direction with positive acceleration

  

  v-t graph of motion in positive direction with negative acceleration

• 

  v-t graph of motion in negative direction with negative acceleration

  

  v-t graph of motion of an object with negative acceleration that changes direction at time t₁.

  

  PROBLEM-1

  Draw v-t graph from the given x-t graph.

  

  Solution

  

  PROBLEM-2

  • Velocity-time graph of a ball thrown vertically upwards with an initial velocity is shown in figure.

  

  1. What is the magnitude of initial velocity of the ball?
  2. Calculate the distance travelled by the ball during 20 s, from the graph.
  3. Calculate the acceleration of the ball from the graph

  Solution 

• 

• ACCELERATION –TIME GRAPH
  

• A graph with acceleration along Y –axis and time along X-axis.

• Area under acceleration – time graph gives velocity.

• PROBLEM 3

  • The graph shows the velocity – time graph of a moving body in a one dimensional motion. Draw the corresponding acceleration – time graph

  

• Solution

• 
  

• 

  PROBLEM -4

  Draw acceleration –time graph from the velocity-time graph given below.

  

  Solution

  

  PROBLEM -5

  which of these cannot possibly represent one-dimensional motion of a particle.

   

  Solution

1. No – because a particle cannot have two positions at the same instant of time.
2. No – because particle can never have two values of velocities at the same instant of time.
3. No- speed cannot be negative
4. No – total path length cannot decrease with time.

5. NATURE OF GRAPHS IN A NUTSHELL

   x-t Graph	v-t Graph	a-t Graph

    

   KINEMATIC EQUATIONS FOR UNIFORMLY ACCELERATED MOTION

   velocity–time graph of an object moving with uniform acceleration and with initial velocity v0 

   Velocity – Time Relation

   • We have

   

   

   • Where v= final velocity, a = acceleration v0 =initial velocity

   at = v − v0

   Or v = v0 + at

   Displacement-Time Relation

   • We know , area under v-t graph = Displacement
   • Thus, the displacement at any time interval 0 and t, is given y

    

   • 
   • But v − v0 = at
   • Thus

   

   • Therefore

   

   • If xo is the initial displacement

   

   Velocity –Displacement Relation

   • We have

   

   • Thus

   Displacement = Average velocity x time

   • Therefore

   

   • But

   

   • Thus

   

   Therefore

   v2 = v02 + 2ax

   • If xo is the initial displacement

   v2 = v02 + 2a (x – x0 )

   Thus the equation of motion are

   

   PROBLEM 1

   • A ball is thrown vertically upwards with a velocity of 20 m s–1 from the top of a multi-storey building. The height of the point from where the ball is thrown is 25.0 m from the ground.

   a) How high will the ball rise ?

   b) how long will it be before the ball hits the

   ground? Take g = 10 m s–2

   Solution

   

   a) Given v0 = +20 m/s , a = -g =10 m/s , v = 0

   • Using the equation

   v2 = v02 + 2a(y – y0)

6. • We get

   (y − y0 ) = 20m

   b). We have y0 = 25 m, y = 0 m, vo = 20 m /s, a = –10m /s2,

   • Using the equation

   

   • Solving this quadratic equation we get, t=5s.

   MOTION OF AN OBJECT UNDER FREE FALL

   • A body falling under the influence of acceleration due to gravity alone is called free fall (air resistance neglected)
   • If the height through which the object falls is small compared to the earth’s radius, g can be taken to be constant, equal to 9.8 m s–2.
   • Free fall is an example of motion with uniform acceleration.
   • Since the acceleration due to gravity is always downward, it is in the negative direction.
   • Acceleration due to gravity = – g = – 9.8ms⁻².

   Equations of motion of a freely falling body

Since the freely falling bodies fall with uniformly accelerated motion, the three equations of motion derived earlier for bodies under uniform acceleration can be applied to the motion of freely falling bodies. For freely falling bodies, the acceleration due to gravity is ‘g’, so we replace the acceleration ‘a’ of the equations by ‘g’ and since the vertical distance of the freely falling bodies is known as height ‘h’, we replace the distance ‘s’ in our equations by the height ‘h’.                         This gives us the following modified equations for the motion of freely falling bodies.


• 
  
  

These modified equations are required  to solve numerical problems. We must remember the following important points for the motion of freely falling bodies.

1. When a body is dropped freely from a height, its initial velocity ‘u’ becomes zero
2. When a body is thrown vertically upwards, its final velocity ‘v’ becomes zero
3. The time taken by body to rise to the highest point is equal to the time it takes to fall from the same height.
4. The distance travelled by a freely falling body is directly proportional to the square of time of fall.

Graphs Related to Freely falling Bodies

1. Acceleration –Time graph of a freely falling body

   

   Velocity – Time graph of a freely falling body

   

   Position –Time graph of a freely falling body

   

   Galileo’s law of odd numbers

   • The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [namely, 1: 3: 5: 7……]

   Proof

   • Divide time interval of motion into equal intervals
   • The distance travelled is found out using

   

   

   • Thus ratio of distances is found to be 1:3:5:7:…..

   STOPPING DISTANCE OF VEHICLES

   • When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance.
   • Stopping distance is an important factor considered in setting speed limits, for example, in school zones
   • Stopping distance depends on the initial velocity (v0) and the braking capacity, or deceleration (–a) that is caused by the braking.

   Equation for Stopping Distance

   • Let the distance travelled by the vehicle before it stops be, d.
   • Substituting v=0 , x=d and acceleration = – a in the equation
   • Thus, stopping distance ,
   • Thus stopping distance is proportional to square initial velocity.

   REACTION TIME

   • Reaction time is the time a person takes to observe, think and act.

   

   • Dropping a ruler the reaction time can be calculated using the formula

   

   • Where d is the distance moved before reaction.

   RELATIVE VELOCITY

   • It is the velocity measured whenever there is a relative motion between objects.
   • The velocity of object B relative to object A is

   vBA = vB – vA

   • Similarly, velocity of object A relative to object B is:

   vAB =vA – vB

2. • Thus, vBA = -vAB

   Special cases:-

   • If vB = vA, – relative velocity vAB or vBA is zero
   • If vA > vB, vB – vA is negative, thus, object A overtakes object B at this time

   Position-time graphs of two objects with equal velocities

   

   Position-time graphs of two objects with unequal velocities

3. 

   Position-time graphs of two objects with velocities in opposite directions

   

4. PROBLEM

   Two parallel rail tracks run north south. Train A moves north with a speed of 54 km h–1, and train B moves south with a speed of 90 km h–1. What is the

   1. Velocity of B with respect to A ?,
   2. b) Velocity of ground with respect to B ?
   3. c) velocity of a monkey running on the roof of the train A against its motion (with a velocity of 18 km h–1 with respect to the train A) as observed by a man standing on the ground ?

   Solution

   

   

   

Revision at a glance :

Motion:

If an object changes its position with respect to its surroundings with time, then it is called in motion.

Rest

If an object does not change its position with respect to its surroundings with time, then it is called at rest.

[Rest and motion are relative states. It means an object which is at rest in one frame of reference can be in motion in another frame of reference at the same time.]

Point Mass Object An object can be considered as a point mass object, if the distance travelled by it in motion is very large in comparison to its dimensions.

Types of Motion

1. One Dimensional Motion

If only one out of three coordinates specifying the position of the object changes with respect to time, then the motion is called one dimensional motion.

For instance, motion of a block in a straight line motion of a train along a straight track a man walking on a level and narrow road and object falling under gravity etc.

2. Two Dimensional Motion

If only two out of three coordinates specifying the position of the object changes with respect to time, then the motion is called two dimensional motion.

A circular motion is an instance of two dimensional motion.

3. Three Dimensional Motion

If all the three coordinates specifying the position of the object changes with respect to time, then the motion is called three dimensional motion.

A few instances of three dimension are flying bird, a flying kite, a flying aeroplane, the random motion of gas molecule etc.

Distance

The length of the actual path traversed by an object is called the distance.

It is a scalar quantity and it can never be zero or negative during the motion of an object.

Its unit is metre.

Displacement

The shortest distance between the initial and final positions of any object during motion is called displacement. The displacement of an object in a given time can be positive, zero or negative.

It is a vector quantity.

Its unit is metre.

Speed

The time rate of change of position of the object in any direction is called speed of the object.

Speed (v) = Distance travelled (s) / Time taken (t)

Its unit is m/s.

It is a scalar quantity.

Its dimensional formula is [M^oT^-1].

Uniform Speed

If an object covers equal distances in equal intervals of time, then its speed is called uniform speed.

Non-uniform or Variable Speed

If an object covers unequal distances in equal intervals of time, then its speed is called non-uniform or variable speed.

Average Speed

The ratio of the total distance travelled by the object to the total time taken is called average speed of the object.

Average speed = Total distanced travelled / Total time taken

If a particle travels distances s₁, s₂, s₃ , … with speeds v₁, v₂, v₃, …, then

Average speed = s₁ + s₂ + s₃ + ….. / (s₁ / v₁ + s₂ / v₂ + s₃ / v₃ + …..)

If particle travels equal distances (s₁ = s₂ = s) with velocities v₁ and v₂, then

Average speed = 2 v₁ v₂ / (v₁ + v₂)

If a particle travels with speeds v₁, v₂, v₃, …, during time intervals t₁, t₂, t₃,…, then

Average speed = v₁t₁ + v₂t₂ + v₃t₃ +… / t₁ + t₂ + t₃ +….

If particle travels with speeds v₁, and v₂ for equal time intervals, i.e., t₁ = t₂ = t₃, then

Average speed = v₁ + v₂ / 2

When a body travels equal distance with speeds V₁ and V₂, the average speed (v) is the harmonic mean of two speeds,

2 / v = 1 / v₁ + 1 / v₂

Instantaneous Speed

When an object is travelling with variable speed, then its speed at a given instant of time is called its instantaneous speed.

 

Velocity

The rate of change of displacement of an object in a particular direction is called its velocity.

Velocity = Displacement / Time taken

Its unit is m/s.

Its dimensional formula is [M^oLT^-1].

It is a vector quantity, as it has both, the magnitude and direction.

The velocity of an object can be positive, zero and negative.

Uniform Velocity

If an object undergoes equal displacements in equal intervals of time, then it is said to be moving with a uniform velocity.

Non-uniform or Variable Velocity

If an object undergoes unequal displacements in equal intervals of time, then it is said to be moving with a non-uniform or variable velocity.

Relative Velocity

Relative velocity of one object with respect to another object is the time rate of change of relative position of one object with respect to another object.

Relative velocity of object A with respect to object B

V_AB = V_A – V_B

When two objects are moving in the same direction, then

When two objects are moving in opposite direction, then

When two objects are moving at an angle, then

and tan β = v_B sin θ / v_A – v_B cos θ

Average Velocity

The ratio of the total displacement to the total time taken is called average velocity.

Average velocity = Total displacement / Total time taken

Acceleration

The time rate of change of velocity is called acceleration.

Acceleration (a) = Change in velocity (Δv) / Time interval (Δt)

Its unit is m/s²

Its dimensional formula is [M^oLT^-2].

It is a vector quantity.

Acceleration can be positive, zero or negative. Positive acceleration means velocity increasing with time, zero acceleration means velocity is uniform while negative acceleration (retardation) means velocity is decreasing with time.

If a particle is accelerated for a time t₁ with acceleration a₁ and for a time t₂ with acceleration a₂, then average acceleration

a_av = a₁t₁ + a₂t₂ / t₁ + t₂

Different Graphs of Motion

Displacement – Time Graph

Note Slope of displacement-time graph gives average velocity.

Velocity – Time Graph

Note Slope of velocity-time graph gives average acceleration.

Acceleration – Time

Equations of Uniformly Accelerated Motion

If a body starts with velocity (u) and after time t its velocity changes to v, if the uniform acceleration is a and the distance travelled in time t in s, then the following relations are obtained, which are called equations of uniformly accelerated motion.

(i) v = u + at

(ii) s = ut + at²

(iii) v² = u² + 2as

(iv) Distance travelled in nth second.

S_n = u + a / 2(2n – 1)

If a body moves with uniform acceleration and velocity changes from u to v in a time interval, then the velocity at the mid point of its path

√u² + v² / 2

Motion Under Gravity

If an object is falling freely (u = 0) under gravity, then equations of motion

(i) v = u + gt

(ii) h = ut + gt²

(iii) V² = u² + 2gh

Note If an object is thrown upward then g is replaced by – g in above three equations.

It thus follows that

(i) Time taken to reach maximum height

t_A = u / g = √2h / g

(ii) Maximum height reached by the body

h_max = u² / 2g

(iii) A ball is dropped from a building of height h and it reaches after t seconds on earth. From the same building if two ball are thrown (one upwards and other downwards) with the same velocity u and they reach the earth surface after t, and t2 seconds respectively, then

t = √t₁t₂

(iv) When a body is dropped freely from the top of the tower and another body is projected horizontally from the same point, both will reach the ground at the same time.

Question 3. 1. In which of the following examples of motion, can the body be considered approximately a point object.
(a) A railway carriage moving without jerks between two stations.
(b) A monkey sitting on top of a man cycling smoothly on a circular track.
(c) A spinning cricket ball that turns sharply on hitting the ground.
(d) A tumbling beaker that has slipped off the edge of table.
Answer:  (a) The railway carriage moving without jerks between two stations, so the distance
between two stations is considered to be large as compared to the size of the train. Therefore the train is considered as a point object.
(b) The monkey may be considered as point object because value of distance covered on
a circular track is much greater.
(c) As turning of ball is not smooth, thus the distance covered by ball is not large in the reasonable time. Therefore ball cannot be considered as point object.
(d) Again a tumbling beaker slipped off the edge of a table cannot be considered as a point object because distance covered is not much larger.

Question 3. 2. The position-time (x -1) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. Choose the correct entries in the brackets below:
(a) (A/B) lives closer to the school than (B/A).
(b) (A/B) starts from the school earlier than (B/A).
(c) (A/B) walks faster than (B/A).
(d) A and B reach home at the (same/different) time.
(e) (A/B) overtakes (B/A) on the road (once/twice).

Answer: (a) A lives closer to school than B, because B has to cover higher distances [OP < OQ],
(b) A starts earlier for school than B, because t = 0 for A but for B, t has some finite time.
(c) As slope of B is greater than that of A, thus B walks faster than A.
(d) A and B reach home at the same time.
(e) At the point of intersection (i.e., X), B overtakes A on the roads once

Question 3. 3. A woman starts from her home at 9.00 am, walks with a speed of 5 km h-1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 2.5 km h-1. Choose suitable scales and plot the x-t graph of her motion.
Answer: Distance covered while walking = 2.5 km.
Speed while walking = 5 km/h
Time taken to reach office while walking = (2.5/5 ) h=1/2 h

If O is regarded as the origin for both time and distance, then at t = 9.00 am, x = 0
and at t = 9.30 am, x = 2.5 km
OA is the x-t graph of the motion when the woman walks from her home to office. Her stay in the office from 9.30 am to 5.00 pm is represented, by the straight line AB in the graph.
Now, time taken to return home by an auto = 2.5/5 h =1/10 h =6 minute
So, at t = 5.06 pm, x = 0
This motion is represented by the straight line BC in the graph. While drawing the x-t graph, the scales chosen are as under:
Along time-axis, one division equals 1 hour.
Along positive-axis, one division equals 0.5 km.

Question 3. 4. A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
Answer:  Since the man steadily moves forward as the time progresses so the following graph will represent his motion till he covers 13 m. In 5 s he moves through a distance of 5 m and then in next 3 s comes back by 3 m.  

Thus in 8 s he covers only 2 m, as shown in the graph he would fall in the pit in 37 s.

As pointed out earlier, the man covers 2 m in 8 s so, he will cover 8 m in 32 s. But at the end in 5 s he would cover another 5 m i.e., 32 s + 5 s = 37 s, he would cover 8 m + 5 m = 13 m. Thus, he would fall in the pit in 37th second.

Question 3. 5. A jet airplane travelling at the speed of 500 km h-1 ejects its products of combustion at the speed of 1500 km h-1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?
Answer: Velocity of jet airplane w.r.t observer on ground = 500 km/h.
If Vj and v0 represent the velocities of jet and observer respectively, then vj – vo = 500 km h-1
Similarly, if vc represents the velocity of the combustion products w.r.t jet plane, then vc – vg = -1500 km/h
The negative sign indicates that the combustion products move in a direction opposite to that of jet.
Speed of combustion products w.r.t. observer
= vc – u0 = (vc – vj) + (vj – v0) = (-1500 + 500) km h-1 = -1000 km h-1.

Question 3. 6. A car moving along a straight highway with speed of 126 km h-1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?
Answer:

Question 3. 7. Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 71 km h-1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 ms-1. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?

Answer: Here length of train A = length of train B = l = 400 m. As speed of both trains u = 72 km h-1 = 20 ms-1 in same direction, hence their relative velocity uBA = 0.
Let initial distance between the two trains be ‘S’ then train B covers the distance (S + 11) = (S + 800) m in time t = 50 s when accelerated with a uniform acceleration a = 1 m/s2.

and initial distance between guard of train B from driver of train A = 450 + 800 = 1250 m.

Question 3. 8. On a two-lane road, car A is travelling with a speed of 36 km h-1. Two cars B and C approach car A in opposite directions with a speed of 54 km h-1 each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?
Answer:

Question 3. 9. Two towns A and B are connected by regular bus service with a bus leaving in either direction every T minute. A man cycling with a speed of 20 km h-1 in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road?
Answer:  Let vb be the speed of each bus. Let vc be the speed of cyclist.
Relative velocity of the buses plying in the direction of motion of cyclist is vb – vc .
The buses playing in the direction of motion of the cyclist go past him after every
18 minute i.e.18/20 h.

Question 3. 10. A player throws a ball upwards with an initial speed of 29.4 ms-1.
(a) What is the direction of acceleration during the upward motion of the ball?
(b) What are the velocity and acceleration of the ball at the highest point of its motion?
(c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player’s hands? (Take g = 9.8 m s-2 and neglect air resistance).
Answer: (a) The direction of acceleration during the upward motion of the ball is vertically downward.
(b) At the highest point, velocity of ball is zero but acceleration (g = 9.8 ms-2) in vertically downward direction.
(c) If we consider highest point of ball motion as x = 0, t = 0 and vertically downward direction to be positive direction of x-axis, then
(i) during upward motion of ball before reaching the highest point position (as well as displacement) x = positive, velocity v = negative and acceleration a = g = positive.
(ii) during the downward motion of ball after reaching the highest point, x, v and a = g all the three quantities are positive.
(d) During upward motion

Question 3. 11. Read each statement below carefully and state with reasons and examples, if it is true or false; A particle in one-dimensional motion
(a) with zero speed at an instant may have non-zero acceleration at that instant.
(b) with, zero speed may have non-zero velocity.
(c) with constant speed must have zero acceleration,
(d) with positive value of acceleration must be speeding up.
Answer:  (a) True. Consider a ball thrown up. At the highest point, speed is zero but the acceleration is non-zero.
(b) False. If a particle has non-zero velocity, it must have speed.
(c) True. If the particle rebounds instantly with the same speed, it implies infinite acceleration which is physically impossible.
(d) False. True only when the chosen position direction is along the direction of motion.

Question 3. 12. A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t =0 to 12 s.
Answer:

Question 3. 13. Explain clearly, with examples, the distinction between:
(a) Magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;
(b) Magnitude of average velocity over an interval of time, and the average speed over the same interval. (Average speed of a particle over an interval of time is defined as the total path length divided by the time interval). Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true? [For simplicity, consider one dimensional motion only],
Answer: (a) Suppose a particle goes from point A to B along a straight path and returns to A along the same path. The magnitude of the displacement of the particle is zero, because the particle has returned to its initial position. The total length of path covered by the particle is AB + BA = AB + AB = 2 AB. Thus, the second quantity is greater than the first,
(b) Suppose, in the above example, the particle takes time t to cover the whole journey. Then, the magnitude of the average velocity of the particle over time-interval t is = Magnitude of displacement /Time-interval =0/t =0
While the average speed of the particle over the same time-interval is =Total path length /Time-interval= 2 AB /t
Again, the second quantity (average speed) is greater than the first (magnitude of average velocity).
Note: In both the above cases, the two quantities are equal if the particle moves from one point to another along a straight path in the same direction only.

Question 3.14. A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h-1 .Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h-1 What is the (a)Magnitude of average velocity, and (b)Average speed of the man over the interval of time (i) 0 to 30 min. (ii) 0 to 50 min. (iii) 0 to 40 min? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero!]
Answer:

Question 3. 15. In Exercises 3.13 and 3.14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
Answer: Instantaneous velocity is the velocity of a particle at a particular instant of time. In this case of small interval of time, the magnitude of the displacement is effectively equal to the distance travelled by the particle in the same interval of time. Therefore, there is no distinction between instantaneous velocity and speed.

Question 3. 16. Look at the graphs (a) to (d) Fig. carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

Answer: None of the four graph represent a possible one-dimensional motion. In graphs (a) and (b) motions are definitely two dimensional. Graph (a) represents two positions at the same time which is not possible.
In graph (b) opposite motion is visible at the same time.
The graph (c) is not correct since it shows that the particle has negative speed at a certain instant. Speed is always positive.
In graph (d) path length is shown as increasing as well as decreasing. Path length never decreases.

Question 3. 17. Figure shows the x-t plot of one-dimensional motion of a particle.
Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0? If not, suggest a suitable physical context for this graph.

Answer:  It is not correct to say that the particle moves in a straight line for t < 0 (i.e., -ve) and on a parabolic path for t > 0 (i.e., + ve)
because the x-t graph can not show the path of the particle.
For the graph, a suitable physical context can be the particle thrown from the top of a tower at the instant t =0.

Question 3. 18. A police van moving on a highway with a speed of 30 km h-1 fires a bullet at a thief s car speeding away in the same direction with a speed of 192 km h-1 . If the muzzle speed of the bullet is 150 ms-1 , with what speed does the bullet hit the thief s car? (Note: Obtain that speed which is relevant for damaging the thief s car).
Answer: 

Question 3. 19.  Suggest a suitable physical situation for each of the following graphs:

Answer: (a) A ball at rest on a smooth floor is kicked. It rebounds from a wall with reduced speed and moves to the opposite wall which stops it.
(b) The graph shows that velocity changes again and again with the passage of time and every time losing some speed. Therefore, it may represent a physical situation such as a ball falling freely (after thrown up), on striking the ground rebounds with reduced speed after each hit against the ground.
(c) A uniformly moving cricket ball turned back by hitting it with a bat for a very short time-interval.

Question 3. 20. Figure gives the x-t plot of a particle executing one dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter 14). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s.
Answer: In x-t graph of Fig. showing simple harmonic motion of a particle, the signs of position, velocity and acceleration are as given below.

In S.H.M., acceleration, a – x or a = – kx.
(i) At t = 0.3 s, x < 0 i.e., x is in -ve direction. Moreover, as x is becoming more negative with time, it shows that v is also – ve (i.e., v < 0). However, a = -kx will be +ve (a > 0).
(ii) At t = 1.2 s, x > 0, v > 0 and a < 0.
(iii) At t = -1.2 s, x < 0, but here on increasing the time t, value of x becomes less negative.
It means that v is +ve (i.e., v > 0). Again a = – kx will be positive (i.e., a > 0).

Question 3. 21. Figure gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval.

Answer: Greater in 3, least in 2; v > 0 in 1 and 2, v < 0 in interval 3.

Question 3. 22. Figure gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at points A, B, C and D?

Answer: The acceleration is greatest in magnitude in interval 2 as the change in speed in the same time is maximum in this interval.
The average speed is greatest in interval 3 (peak D is at maximum on speed axis).
The sign of v and a in the three intervals are:
v > 0 in 1, 2 and 3; a > 0 in 1
a < 0 in 2, a=0 in 3.
acceleration is zero at A,B,C and D.

Question 3. 23. A three – wheeler starts from rest , accelerates uniformly with 1 m s-2 on a straight road for 10 s, and then moves with uniform velocity .plot the distance covered by the vehicle during the n th second (n=1,2,3……..) versus n. what do you expect this plot to be during accelerated motion: a straight line or a parabola?
Answer: 

Question 3. 24. A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m s-1. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m s-1 and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?
Answer:  When either the lift is at rest or the lift is moving either vertically upward or downward with a constant speed, we can apply three simple kinematic motion equations presuming a = ± g (as the case may be).In present case u = 49 ms-1 (upward) a = g = 9.8 ms-2 (downward)
If the ball returns to boy’s hands after a time t, then displacement of ball relative to boy
is zero i.e., s = 0. Hence, using equation s = ut + 1/2 at 2, we have

Question 3. 25. On a long horizontally moving belt (Fig.), a child runs to and fro with n speed 9 km h-1 (with respect to the belt) between his father and mother located 50 a part on the moving belt. The belt moves with a speed of 4 km h-1 . For an observe a stationary platform outside, what is the
(a) Speed of the child running in the direction of motion of the belt?
(b) Speed of the child running opposite to the direction of motion of the belt?
(c) Time taken by the child in (a) and (b)?
Which of the answers alter if motion is viewed by one of the parents?

Answer: Speed of child with respect to belt = 9 km h-1 Speed of belt = 4 km h-1
(a) When the child runs in the direction of motion of the belt, then speed of child w.r.t. stationary observer = (9 + 4) km h-1 = 13 km h-1.
(b) When the child runs opposite to the direction of motion of the belt, then speed of child w.r.t. stationary observer = (9 – 4) km h-1 = 5 km h-1
(c) Speed of child w.r.t . either parent=9 km h-1
Distance to be covered = 50 m = 0.05 km 0.05 km
Time=0.05 km/9k h-1=0.0056 h =20 S
If the motion is viewed by one of the parents, then the answers to (a) and (b)are altered but answer to (c) remains unaltered.

Question 3. 26. Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 ms-1 and 30 ms-1. Verify that the graph shown in Fig. correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10 ms-2. Give the equations for the linear and curved parts of the plot.

Answer: For first stone,
x (0) = 200 m, v (0) = 15 ms-1, a = -10 ms-2
x1 (t) = x (0) + v (0) t + 1/2 a t2
x1 (t) = 200 + 15t – 5t2
When the first stone hits the ground, x1 (t) = 0
– 5t2 + 15t+ 200 = 0 On simplification, t = 8 s
For second stone, x (0) = 200 m, v (0) = 30 ms-1, a = -10 ms-2
x1 (t) = 200 + 30t – 5t2
When this stone hits the ground, x1(t) = 0 .-. -5t2 + 30t + 200 = 0
Relative position of second stone w.r.t. first is given by x2 (t) – x1 (t) = 15t
Since there is a linear relationship between x2(t) – x1 (t) and t, therefore the graph is a straight line.
For maximum separation, t = 8 s So maximum separation is 120 m
After 8 second, only the second stone would be in motion. So, the graph is in accordance with the quadratic equation.

Question 3. 27. The speed-time graph of a particle moving along a fixed direction is shown in Fig. Obtain the distance traversed by the particle between
(a) t = 0 s to 10 s. (b) t = 2 s to 6 s.
What is the average speed of the particle over the intervals in (a) and (b)?

Answer: 

Question 3. 28.  The velocity-time graph of a particle in one-dimensional motion is shown below. Which of the following formula are correct for describing the motion of the particle over the time interval from t1 to t2?

Answer: (c),(d),(f).
As it is evident from the shape of v-t graph that acceleration of the particle is not uniform between time intervals t1 and t2. (since the given v-t graph is not straight). The equations (a), (b) and (e) represent uniform acceleration.

Motion in a Straight Line   MCQs  

Multiple Choice Questions

Question 1.
The displacement-time graph of a moving object is a straight line. Then,
(a) its acceleration may be uniform
(b) its velocity may be uniform
(c) its acceleration may be variable
(d) both its velocity and acceleration may be uniform

Answer

---

Question 2.
If the displacement of an object is zero, then what can we say about its distance covered?
(a) It is negative
(b) It is must be zero
(c) It cannot be zero
(d) It may or may not be zero

Answer

---

Question 3.
Which of the following changes when a particle is moving with uniform velocity?
(a) Speed
(6) Velocity
(c) Acceleration
(d) Position vector

Answer

---

Question 4.
The distance travelled by an object is directly proportional to the time taken. Its acceleration
(a) increases
(b) decreases
(c) becomes zero
(d) remains constant

Answer

---

Question 5.
The distance travelled by an object is directly proportional to the time taken. Its speed
(a) increases
(b) decreases
(c) becomes zero
(d) remains constant

Answer

---

Question 6.
A particle is moving with a constant speed along straight line path. A force is not required to
(a) change its direction
(b) increase its speed
(c) decrease its momentum
(d) keep it moving with uniform velocity

Answer

---

Question 7.
If the velocity-time graph of an object is a straight line sloping downwards, the body has
(a) zero acceleration
(b) positive acceleration
(c) constant acceleration
(d) negative acceleration

Answer

---

Question 8.
When a body is dropped from a tower, then there is an increase in its
(a) mass
(b) velocity
(c) acceleration
(d) potential energy

Answer

---

Question 9.
If s represents distance and S represents displacement, then S/s|S|s is.
(a) > 1
(b) < 1
(c) = 1
(d) ≤ 1

Answer

---

Question 10.
The velocity time graph of motion of an object starting from rest with uniform acceleration is a straight line
(a) parallel to time axis
(b) parallel to velocity axis
(c) passing through origin
(d) none of the above

Answer

---

Question 11.
If the displacement-time graph of an object is parallel to the time-axis, then it represents that the object is :
(a) at rest
(b) in uniform motion
(c) in acceleration motion
(d) none of the above

Answer

---

Question 12.
If the displacement of a given body is found to be directly proportional to the cube of the time elapsed, then the magnitude of the acceleration of the body is
(a) zero
(b) constant but not zero
(c) increasing with time
(d) decreasing with time

Answer

---

Question 13.
The acceleration of a moving object can be found from
(a) area under displacement-time graph
(b) slope of displacement-time graph
(c) area under velocity-time graph
(d) slope of velocity-time graph

Answer

---

Question 14.
The total vertical distance covered by a freely falling body in a given time is directly proportional to
(a) time
(b) square of time
(c) square of acceleration due to gravity
(d) product of the time and acceleration due to gravity

Answer

---

Question 15.
A simple pendulum hangs from the roof of a train. The string is inclined towards the rear of the train. What is the nature of motion of the train?
(a) Uniform
(b) Accelerated
(c) Retarded
(d) At rest

Answer

---

Question 16.
A bucket is placed in the open where the rain is falling vertically. If a wind begins to blow at double the velocity of the rain, how will be the rate of filling of the bucket change?
(a) Remains unchanged
(b) Doubled
(c) Halved
(d) Becomes four times

Answer

---

Question 17.
Two particles start from rest simultaneously and are equally accelerated. Throughout the motion, the relative velocity of one w.r.t. other is :
(а) zero
(b) non-zero and directed parallel to acceleration
(c) non-zero and directed opposite to acceleration
(d) directed perpendicular to the acceleration

Answer

---

Question 18.
The relative velocity of a particle moving with a velocity v w.r.t. itself is
(a) v
(b) -v
(c) zero
(d) none of the above

Answer

---

Question 19.
A stone is thrown upward and it rises to a height of 100 m. The relative velocity of the stone w.r.t. the Earth will be maximum at:
(a) the ground
(b) a height of 50 m
(c) a height of 5 m
(d) the highest point

Answer

---

Question 20.
Two particles are moving with velocities v1 and v2. Their relative velocity is maximum when the angle between their velocities
(a) zero
(b) π2
(c) π
(d) π4

Answer

---

Fill in the blanks

Question 1.
The distances travelled by a body falling freely from rest in the first, second and third seconds are in the ratio …………………

Answer

---

Question 2.
The slope of the velocity-time graph for accelerated motion is …………………

Answer

---

Question 3.
The velocity-time graph of the body at rest is …………………

Answer

---

Question 4.
A moving body is covering distances in proportion to the square of the time along a straight line, its acceleration is …………………

Answer

---

Question 5.
The area under v-t graph for uniform motion gives …………………

Answer

---

Question 6.
If the distance covered by a particle is zero, then its displacement …………………

Answer

---

Question 7.
If the displacement of a particle is zero, then the distance covered by it …………………

Answer

---

Question 8.
A car travels a distance s on a straight road in two hours and then returns to the starting point in the next three hours, its average velocity will be …………………

Answer

---

Question 9.
A particle moves along a straight line path. After sometime it comes to rest. The motion is with constant acceleration whose direction with respect to the direction of velocity is …………………

Answer

---

Question 10.
In a one dimensional motion with constant acceleration, the time rate of change of speed is independent of …………………

Answer

---

True/False Type Questions

Question 1.
Displacement is independent of the choice of origin of the axis.

Answer

---

Question 2.
Displacement may or may not be equal to the distance travelled.

Answer

---

Question 3.
When a particle returns to its starting point, its displacement is zero.

Answer

---

Question 4.
A positive acceleration always corresponds to the speeding up and a negative acceleration always corresponds to the speeding down.

Answer

---

Question 5.
Speed can never be negative.

Answer

---

Question 6.
If the velocity of a particle is zero at an instant, its acceleration should also be zero at that instant.

Answer

---

Question 7.
When the particle returns to the starting point, its average velocity is zero but average speed is not zero.

Answer

---

Question 8.
Displacement does not tell the nature of the actual motion of a particle between the points.

Answer

---

Question 9.
non-zero acceleration without having varying speed.

Answer

---

Question 10.
non-zero acceleration without having vary ing velocity.

Answer

---

Question 11.
varying speed without having varying velocity.

Answer

---

Question 12.
varying velocity without having varying speed.

Answer

---

Question 13.
If the velocity is zero at any instant, the acceleration should also be zero at that instant.

Answer

---

Question 14.
If the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval.

Answer

---

Question 15.
If the position and velocity have opposite signs, the particle is moving towards the origin.

Answer

---

Question 16.
If the velocity and acceleration have opposite signs, the object is slowing down.

Answer

---

Question 17.
It is possible to have a situation in which the speed of the particle is never zero but the average speed in an interval is zero.

Answer

---

Question 18.
It is possible to have a situation in which the speed of a particle is always zero but the average speed is not zero.

Answer

---

Question 19.
The magnitude of the velocity of a particle is equal to its speed.

Answer

---

Question 20.
The magnitude of average velocity in an interval is equal to its average speed in that interval.

Answer

---

Question 21.
square of time

Answer

---

Question 22.
time

Answer

---

Question 23.
square of acceleration due to gravity

Answer

---

Question 24.
product of acceleration due to gravity and time.

Answer

---

Question 25.
Displacement is always positive.

Answer

---

Question 26.
Displacement has both magnitude and direction.

Answer

---

Question 27.
Displacement can be represented geometrically.

Answer

---

Question 28.
Magnitude of displacement is equal to the shortest distance between the initial and final positions of the object.

Answer

---

Question 29.
The motion is always in the same direction.

Answer

---

Question 30.
The motion is along a straight line path.

Answer

---

Question 31.
Average velocity is equal to the instantaneous velocity.

Answer

---

Question 32.
Magnitude of displacement < distance covered.

Answer

---

Question 33.
Its speed is zero.

Answer

---

Question 34.
Its acceleration is zero.

Answer

---

Question 35.
Its acceleration is opposite to the velocity.

Answer

---

Question 36.
Its speed may be variable.

Answer

---

If the distance covered by a particle is zero, then which statement is true/false.
Question 37.
It is negative.

Answer

---

Question 38.
It must be zero.

Answer

---

Question 39.
It cannot be zero.

Answer

---

Question 40.
It may or may not be zero.

Answer

---

Match type Questions:

Column I	Column II
(1) Motion with uniform velocity	(a) parallel to time axis.
(2) Angle between the  instantaneous displacement and acceleration during the retarded motion	(b) distance covered during that motion.
(3) Velocity-time graph of uniform motion	(c) zero
(4) Area under velocity-time graph for a given motion	(d) 180°
(5) Slope of velocity-time graph for motion with uniform velocity

Question 2.
If the displacement of an object is zero, then what can we say about its distance covered?
(a) It is negative
(b) It is must be zero
(c) It cannot be zero
(d) It may or may not be zero

Answer

Question 3.
Which of the following changes when a particle is moving with uniform velocity?
(a) Speed
(6) Velocity
(c) Acceleration
(d) Position vector

Question 4.
The distance travelled by an object is directly proportional to the time taken. Its acceleration
(a) increases
(b) decreases
(c) becomes zero
(d) remains constant

Question 5.
The distance travelled by an object is directly proportional to the time taken. Its speed
(a) increases
(b) decreases
(c) becomes zero
(d) remains constant

Question 6.
A particle is moving with a constant speed along straight line path. A force is not required to
(a) change its direction
(b) increase its speed
(c) decrease its momentum
(d) keep it moving with uniform velocity

Question 7.
If the velocity-time graph of an object is a straight line sloping downwards, the body has
(a) zero acceleration
(b) positive acceleration
(c) constant acceleration
(d) negative acceleration

Question 8.
When a body is dropped from a tower, then there is an increase in its
(a) mass
(b) velocity
(c) acceleration
(d) potential energy

Question 9.
If s represents distance and S represents displacement, then S/s|S|s is.
(a) > 1
(b) < 1
(c) = 1
(d) ≤ 1

Question 10.
The velocity time graph of motion of an object starting from rest with uniform acceleration is a straight line
(a) parallel to time axis
(b) parallel to velocity axis
(c) passing through origin
(d) none of the above

Question 11.
If the displacement-time graph of an object is parallel to the time-axis, then it represents that the object is :
(a) at rest
(b) in uniform motion
(c) in acceleration motion
(d) none of the above

Question 12.
If the displacement of a given body is found to be directly proportional to the cube of the time elapsed, then the magnitude of the acceleration of the body is
(a) zero
(b) constant but not zero
(c) increasing with time
(d) decreasing with time

Question 13.
The acceleration of a moving object can be found from
(a) area under displacement-time graph
(b) slope of displacement-time graph
(c) area under velocity-time graph
(d) slope of velocity-time graph

Question 14.
The total vertical distance covered by a freely falling body in a given time is directly proportional to
(a) time
(b) square of time
(c) square of acceleration due to gravity
(d) product of the time and acceleration due to gravity

Question 15.
A simple pendulum hangs from the roof of a train. The string is inclined towards the rear of the train. What is the nature of motion of the train?
(a) Uniform
(b) Accelerated
(c) Retarded
(d) At rest

Question 16.
A bucket is placed in the open where the rain is falling vertically. If a wind begins to blow at double the velocity of the rain, how will be the rate of filling of the bucket change?
(a) Remains unchanged
(b) Doubled
(c) Halved
(d) Becomes four times

Question 17.
Two particles start from rest simultaneously and are equally accelerated. Throughout the motion, the relative velocity of one w.r.t. other is :
(а) zero
(b) non-zero and directed parallel to acceleration
(c) non-zero and directed opposite to acceleration
(d) directed perpendicular to the acceleration

Question 18.
The relative velocity of a particle moving with a velocity v w.r.t. itself is
(a) v
(b) -v
(c) zero
(d) none of the above

Question 19.
A stone is thrown upward and it rises to a height of 100 m. The relative velocity of the stone w.r.t. the Earth will be maximum at:
(a) the ground
(b) a height of 50 m
(c) a height of 5 m
(d) the highest point

Question 20.
Two particles are moving with velocities v1 and v2. Their relative velocity is maximum when the angle between their velocities
(a) zero
(b) π2
(c) π
(d) 

Fill in the blanks

Question 1.
The distances travelled by a body falling freely from rest in the first, second and third seconds are in the ratio …………………

 Question 2.

The slope of the velocity-time graph for accelerated motion is ………………

Question 3.
The velocity-time graph of the body at rest is …………………

Question 4.
A moving body is covering distances in proportion to the square of the time along a straight line, its acceleration is …………………

Question 5.
The area under v-t graph for uniform motion gives …………

Question 6.
If the distance covered by a particle is zero, then its displacement ………

Question 7.
If the displacement of a particle is zero, then the distance covered by it ………………

Question 8.
A car travels a distance s on a straight road in two hours and then returns to the starting point in the next three hours, its average velocity will be …………………

Question 9.
A particle moves along a straight line path. After sometime it comes to rest. The motion is with constant acceleration whose direction with respect to the direction of velocity is …………………

Question 9.
A particle moves along a straight line path. After sometime it comes to rest. The motion is with constant acceleration whose direction with respect to the direction of velocity is …………………

Match type Questions:

Column I	Column II
(1) Motion with uniform velocity	(a) parallel to time axis.
(2) Angle between the  instantaneous displacement and acceleration during the retarded motion	(b) distance covered during that motion.
(3) Velocity-time graph of uniform motion	(c) zero
(4) Area under velocity-time graph for a given motion	(d) 180°
(5) Slope of velocity-time graph for motion with uniform velocity	(e) Average speed = average velocit

Thank you for studying till the end.

---

Question 2.
If the displacement of an object is zero, then what can we say about its distance covered?
(a) It is negative
(b) It is must be zero
(c) It cannot be zero
(d) It may or may not be zero

Answer

---

Question 3.
Which of the following changes when a particle is moving with uniform velocity?
(a) Speed
(6) Velocity
(c) Acceleration
(d) Position vector

Answer

---

Question 4.
The distance travelled by an object is directly proportional to the time taken. Its acceleration
(a) increases
(b) decreases
(c) becomes zero
(d) remains constant

Answer

---

Question 5.
The distance travelled by an object is directly proportional to the time taken. Its speed
(a) increases
(b) decreases
(c) becomes zero
(d) remains constant

Answer

---

Question 6.
A particle is moving with a constant speed along straight line path. A force is not required to
(a) change its direction
(b) increase its speed
(c) decrease its momentum
(d) keep it moving with uniform velocity

Answer

---

Question 7.
If the velocity-time graph of an object is a straight line sloping downwards, the body has
(a) zero acceleration
(b) positive acceleration
(c) constant acceleration
(d) negative acceleration

Answer

---

Question 8.
When a body is dropped from a tower, then there is an increase in its
(a) mass
(b) velocity
(c) acceleration
(d) potential energy

Answer

---

Question 9.
If s represents distance and S represents displacement, then S/s|S|s is.
(a) > 1
(b) < 1
(c) = 1
(d) ≤ 1

Answer

---

Question 10.
The velocity time graph of motion of an object starting from rest with uniform acceleration is a straight line
(a) parallel to time axis
(b) parallel to velocity axis
(c) passing through origin
(d) none of the above

Answer

---

Question 11.
If the displacement-time graph of an object is parallel to the time-axis, then it represents that the object is :
(a) at rest
(b) in uniform motion
(c) in acceleration motion
(d) none of the above

Answer

---

Question 12.
If the displacement of a given body is found to be directly proportional to the cube of the time elapsed, then the magnitude of the acceleration of the body is
(a) zero
(b) constant but not zero
(c) increasing with time
(d) decreasing with time

Answer

---

Question 13.
The acceleration of a moving object can be found from
(a) area under displacement-time graph
(b) slope of displacement-time graph
(c) area under velocity-time graph
(d) slope of velocity-time graph

Answer

---

Question 14.
The total vertical distance covered by a freely falling body in a given time is directly proportional to
(a) time
(b) square of time
(c) square of acceleration due to gravity
(d) product of the time and acceleration due to gravity

Answer

---

Question 15.
A simple pendulum hangs from the roof of a train. The string is inclined towards the rear of the train. What is the nature of motion of the train?
(a) Uniform
(b) Accelerated
(c) Retarded
(d) At rest

Answer

---

Question 16.
A bucket is placed in the open where the rain is falling vertically. If a wind begins to blow at double the velocity of the rain, how will be the rate of filling of the bucket change?
(a) Remains unchanged
(b) Doubled
(c) Halved
(d) Becomes four times

Answer

---

Question 17.
Two particles start from rest simultaneously and are equally accelerated. Throughout the motion, the relative velocity of one w.r.t. other is :
(а) zero
(b) non-zero and directed parallel to acceleration
(c) non-zero and directed opposite to acceleration
(d) directed perpendicular to the acceleration

Answer

---

Question 18.
The relative velocity of a particle moving with a velocity v w.r.t. itself is
(a) v
(b) -v
(c) zero
(d) none of the above

Answer

---

Question 19.
A stone is thrown upward and it rises to a height of 100 m. The relative velocity of the stone w.r.t. the Earth will be maximum at:
(a) the ground
(b) a height of 50 m
(c) a height of 5 m
(d) the highest point

Answer

---

Question 20.
Two particles are moving with velocities v1 and v2. Their relative velocity is maximum when the angle between their velocities
(a) zero
(b) π2
(c) π
(d) π4

Answer

---

Fill in the blanks

Question 1.
The distances travelled by a body falling freely from rest in the first, second and third seconds are in the ratio …………………

Answer

---

Question 2.
The slope of the velocity-time graph for accelerated motion is …………………

Answer

---

Question 3.
The velocity-time graph of the body at rest is …………………

Answer

---

Question 4.
A moving body is covering distances in proportion to the square of the time along a straight line, its acceleration is …………………

Answer

---

Question 5.
The area under v-t graph for uniform motion gives …………………

Answer

---

Question 6.
If the distance covered by a particle is zero, then its displacement …………………

Answer

---

Question 7.
If the displacement of a particle is zero, then the distance covered by it …………………

Answer

---

Question 8.
A car travels a distance s on a straight road in two hours and then returns to the starting point in the next three hours, its average velocity will be …………………

Answer

---

Question 9.
A particle moves along a straight line path. After sometime it comes to rest. The motion is with constant acceleration whose direction with respect to the direction of velocity is …………………

Answer

---

Question 10.
In a one dimensional motion with constant acceleration, the time rate of change of speed is independent of …………………

Answer

---

True/False Type Questions

Question 1.
Displacement is independent of the choice of origin of the axis.

Answer

---

Question 2.
Displacement may or may not be equal to the distance travelled.

Answer

---

Question 3.
When a particle returns to its starting point, its displacement is zero.

Answer

---

Question 4.
A positive acceleration always corresponds to the speeding up and a negative acceleration always corresponds to the speeding down.

Answer

---

Question 5.
Speed can never be negative.

Answer

---

Question 6.
If the velocity of a particle is zero at an instant, its acceleration should also be zero at that instant.

Answer

---

Question 7.
When the particle returns to the starting point, its average velocity is zero but average speed is not zero.

Answer

---

Question 8.
Displacement does not tell the nature of the actual motion of a particle between the points.

Answer

---

Question 9.
non-zero acceleration without having varying speed.

Answer

---

Question 10.
non-zero acceleration without having vary ing velocity.

Answer

---

Question 11.
varying speed without having varying velocity.

Answer

---

Question 12.
varying velocity without having varying speed.

Answer

---

Question 13.
If the velocity is zero at any instant, the acceleration should also be zero at that instant.

Answer

---

Question 14.
If the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval.

Answer

---

Question 15.
If the position and velocity have opposite signs, the particle is moving towards the origin.

Answer

---

Question 16.
If the velocity and acceleration have opposite signs, the object is slowing down.

Answer

---

Question 17.
It is possible to have a situation in which the speed of the particle is never zero but the average speed in an interval is zero.

Answer

---

Question 18.
It is possible to have a situation in which the speed of a particle is always zero but the average speed is not zero.

Answer

---

Question 19.
The magnitude of the velocity of a particle is equal to its speed.

Answer

---

Question 20.
The magnitude of average velocity in an interval is equal to its average speed in that interval.

Answer

---

Question 21.
square of time

Answer

---

Question 22.
time

Answer

---

Question 23.
square of acceleration due to gravity

Answer

---

Question 24.
product of acceleration due to gravity and time.

Answer

---

Question 25.
Displacement is always positive.

Answer

---

Question 26.
Displacement has both magnitude and direction.

Answer

---

Question 27.
Displacement can be represented geometrically.

Answer

---

Question 28.
Magnitude of displacement is equal to the shortest distance between the initial and final positions of the object.

Answer

---

Question 29.
The motion is always in the same direction.

Answer

---

Question 30.
The motion is along a straight line path.

Answer

---

Question 31.
Average velocity is equal to the instantaneous velocity.

Answer

---

Question 32.
Magnitude of displacement < distance covered.

Answer

---

Question 33.
Its speed is zero.

Answer

---

Question 34.
Its acceleration is zero.

Answer

---

Question 35.
Its acceleration is opposite to the velocity.

Answer

---

Question 36.
Its speed may be variable.

Answer

---

If the distance covered by a particle is zero, then which statement is true/false.
Question 37.
It is negative.

Answer

---

Question 38.
It must be zero.

Answer

---

Question 39.
It cannot be zero.

Answer

---

Question 40.
It may or may not be zero.

Answer

---

Match type Questions:

Column I	Column II
(1) Motion with uniform velocity	(a) parallel to time axis.
(2) Angle between the  instantaneous displacement and acceleration during the retarded motion	(b) distance covered during that motion.
(3) Velocity-time graph of uniform motion	(c) zero
(4) Area under velocity-time graph for a given motion	(d) 180°
(5) Slope of velocity-time graph for motion with uniform velocity