- distance and displacement
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- Velocity and speed
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- Instantaneous velocity and speed
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- Acceleration
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- Kinematic equations for uniformly accelerated Motion
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- Free fall acceleration
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- Relative velocity
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- Kinematics Sample Problems and Solutions
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- One dimensional motion problems with solution
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- Motion graphs worksheet with Answer

- Important Questions on Kinematics
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- Projectile motion problems
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- Kinematics and Projectile motion worksheet
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- Motion in one dimension (obj)
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- Motion in one dimension (sub)
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- Motion in one dimension Practice Paper
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- Acceleration worksheet with answers

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 a table.

(a) The size of a carriage is very small as compared to the distance between two stations. Therefore, the carriage can be treated as a point sized object.

(b) The size of a monkey is very small as compared to the size of a circular track. Therefore, the monkey can be considered as a point sized object on the track.

(c) The size of a spinning cricket ball is comparable to the distance through which it turns sharply on hitting the ground. Hence, the cricket ball cannot be considered as a point object.

(d) The size of a beaker is comparable to the height of the table from which it slipped. Hence, the beaker cannot be considered as a point object.

Hence (a) and (b) are correct Solutions

The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown. 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).

(a) As

(b) For

(c) We know that velocity is equal to slope of

(d) It is clear from the given graph that both

(e) The x-t graph intersects only once for A and B. Also, B moves later than

A woman starts from her home at 9.00 am, walks with a speed of 5 km/h 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 25 km/hr. Choose suitable scales and plot the

Speed of the woman = 5 km/h

Distance between her office and home = 2.5 km

Time taken = Distance / Speed

= 2.5 / 5 = 0.5 h = 30 min

So, Women starts at 9:00 AM and reaches office by 9:30 AM

Hence between 9:30 AM and 5:00 PM, she stays in the office which is at distance 2.5 Km from home

It is given that she covers the same distance in the evening by an auto.

Now, speed of the auto = 25 km/h

Time taken = Distance / Speed

= 2.5 / 25 = 1 / 10 = 0.1 h = 6 min

Hence, she starts at 5:00 PM from office and reaches home by 5:06 PM

The suitable

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

We can see from the question

Distance covered with 1 step = 1 m

Time taken = 1 s

Time taken to move first 5 m forward = 5 s

Time taken to move 3 m backward = 3 s

So net distance after 8 secs= 2m

Similarly, we can find the data point for values also

The position of the drunkard taking starting point as position with respect to time is given the below table

T(s) |
0 |
5 |
8 |
13 |
16 |
21 |
24 |
29 |
32 |
37 |

X(m) |
0 |
5 |
2 |
7 |
4 |
9 |
6 |
11 |
8 |
13 |

So, we can see that, drunkard will fall in pit at 37 sec

Net distance covered after 8sec = 5 – 3 = 2 m

Net time taken to cover 2 m = 8 s

So Drunkard covered 8 m in = 4X8= 32 s.

In the next 5 s, the drunkard will cover a distance of 5 m and a total distance of 13 m and falls into the pit.

Net time taken by the drunkard to cover 13 m = 32 + 5 = 37 s

A jet airplane travelling at the speed of 500 km/h ejects its products of combustion at the speed of 1500 km/h relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?

Speed of the jet airplane,

Let v

Let’s take jet direction as positive direction, then

Relative speed of its products of combustion with respect to the plane,

v

The negative sign indicates that the direction of its products of combustion is opposite to the direction of motion of the jet airplane.

A car moving along a straight highway with speed of 126 km/h 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?

Initial velocity of the car,

Final velocity of the car,

Distance covered by the car before coming to rest,

Retardation produced in the car =

From third equation of motion,

v

(0)

a = - 35 × 35

From first equation of motion, time (

v = u + at

t = (v - u) / a = (- 35)

Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km/h in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m/s

Initial velocity,

Time,

Acceleration,

From second equation of motion, distance S

S

= 20 × 50 + 0 = 1000 m

Initial velocity,

Acceleration,

Time,

From second equation of motion, distance S

S

= 20 X 50 + (1/2) × 1 × (50)

Hence, the original distance between the driver of train A and the guard of train B is 2250 - 1000 = 1250m.

On a two-lane road, car A is travelling with a speed of 36 km/h. Two cars B and C approach car A in opposite directions with a speed of 54 km/h 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?

Velocity of car A,

Velocity of car B,

Velocity of car C,

Situation is depicted below

Taking direction of Car A as positive

Hence Relative velocity of car B with respect to car A,

= 15 – 10 = 5 m/s

Relative velocity of car C with respect to car A,

= -15 - 10 = -25 m/s

At a certain instance, both cars B and C are at the same distance from car A i.e.,

Time taken (

Hence, to avoid an accident, car B must cover the same distance in a maximum of 40 s.

From second equation of motion, minimum acceleration (

s = ut + (1/2)at

1000 = 5 × 40 + (1/2) × a × (40)

a = 1600 / 1600 = 1 m/s

Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h

Let

Speed of the cyclist,

Buses are leaving A every T min. So, distance between two buses plying in same direction at any time would be V(T/60) Km

At an instance when one bus passes through the cycle, so another bus would be V(T/60) Km behind it.

Now that another bus must pass the cyclist in 18 min (as buses passes through cyclist every 18 min). Now this can be solved using relative velocity concept

Relative speed of the another bus moving in the direction of the cyclist

=

Time taken =18min=18/60 h

Distance= V(T/60)

Distance=Relative speed X time

(V - 20) × 18

Again, Buses are leaving B every T min. So, distance between two buses plying in same direction at any time would be V(T/60) Km

At an instance when one bus passes through the cycle, so another bus would be V(T/60) Km behind it.

Now that another bus must pass the cyclist in 6 min (as buses passes through cyclist every 6 min)

Relative speed of the bus moving in the opposite direction of the cyclist

= (

Time taken by the bus to go past the cyclist = 6 min = 6 / 60 h

(

From equations (1) and (2), we get

(

2

Substituting the value of

(40 + 20) × 6

T = 360 / 40 = 9 min

A player throws a ball upwards with an initial speed of 29.4 m/s

(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

(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

(a)Irrespective of the direction of the motion of the ball, acceleration (which is acceleration due to gravity) always acts in the downward direction towards the centre of the Earth.

(b)At maximum height, velocity of the ball becomes zero. Acceleration due to gravity at a given place is constant and acts on the ball at all points (including the highest point) with a constant value i.e., 9.8 m/s

(c) During upward motion, the sign of position is positive, sign of velocity is negative, and sign of acceleration is positive. During downward motion, the signs of position, velocity, and acceleration are all positive.

(d)Initial velocity of the ball,

Final velocity of the ball,

Acceleration,

From third equation of motion, height (

v

Now x=0

x

= ((0)

Let t be the time taken to reach the highest point

From First equation of motion

v=u + at

0=-29.4+ 9.8t

t= 3sec

As Time of ascent = Time of descent

Hence, the total time taken by the ball to return to the boy hands = 3 + 3 = 6 s.

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.

(a)

(b)

(c)

(d) False, If the initial velocity of the body is negative and acceleration is positive Then, for all the time before velocity becomes zero, there is slowing down of the particle. Such a case happens when a particle is projected upwards.

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

i)

Ball is dropped from a height,

Initial velocity of the ball,

Acceleration,

Final velocity of the ball =

From second equation of motion, time (

s = ut + (1/2)at

90 = 0 + (1/2) × 9.8 t

t = √

From first equation of motion, final velocity is given as:

= 0 + 9.8 × 4.29 = 42.04 m/s

Here we see that v=at, so velocity varies linearly with downward motion of ball.

This part of motion is represented by first line (1) in the below graph

Now Rebound velocity of the ball,

This is represented the line (2) in the graph. We are assumed negligible time of collision between ball and floor

iii)

Time (

0 = 37.84 + (– 9.8)

Total time taken by the ball =

This is represented by line (3) in the graph

iv)

As the time of ascent is equal to the time of descent, the ball takes 3.86 s to strike back on the floor for the second time.

The velocity with which the ball rebounds from the floor = 9 × 37.84 / 10 = 34.05 m/s

Total time taken by the ball for second rebound = 8.15 + 3.86 = 12.01 s

This is represented by line (4) in the graph

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].

(a) The magnitude of displacement over an interval of time is the shortest distance (which is a straight line) between the initial and final positions of the particle.

The total path length of a particle is the actual path length covered by the particle in each interval of time.

This can be understood with the below example

Suppose a particle moves from point A to point B and then, comes back to a point, C taking a total time t, as shown below.

Now the magnitude of displacement of the particle is given by = AC.

And total path length is given b = AB + BC

So here Total Path Length > displacement

It is also important to note that the magnitude of displacement can never be greater than the total path length. However, in some cases, both quantities are equal.

Let instead of coming back to C from B, particle moves to position D

Then Magnitude of displacement = AD

Total path length =AB +BD

So, both the quantities are equal now

So, both quantities will be equal if the particle continues to move along a straight line.

(b) Magnitude of average velocity = Magnitude of displacement / Time interval

For the given particle,

Average velocity = AC /

Average speed = Total path length / Time interval

= (AB + BC) /

Since (AB + BC) > AC, average speed is greater than the magnitude of average velocity.

Now for the second case

Average velocity=AD/t

Average speed= (AB +BD)/t

So, both the quantities are equal

So, both quantities will be equal if the particle continues to move along a straight line.

A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km/h. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km/h. 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!]

Time taken by the man to reach the market from home,

Time taken by the man to reach home from the market,

Total time taken in the whole journey = 30 + 20 = 50 min

(i) 0 to 30 min

Average velocity = Displacement/Time = 2.5

Average speed = Distance/Time = 2.5

(ii) 0 to 50 min

Time = 50 min

Net displacement = 0

Total distance = 2.5 + 2.5 = 5 km

Average velocity = Displacement / Time = 0

Average speed = Distance / Time = 5/ (5/6) = 6 km/h

(iii) 0 to 40 min

Speed of the man = 7.5 km/h

Distance travelled in first 30 min = 2.5 km

Distance travelled by the man (from market to home) in the next 10 min

= 7.5 × 10/60 = 1.25 km

Net displacement = 2.5 – 1.25 = 1.25 km

Total distance travelled = 2.5 + 1.25 = 3.75 km

Average velocity = Displacement

Average speed = Distance

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?

Instantaneous velocity is given by the first derivative of distance with respect to time i.e.,

v = dx

Here, the time interval

Therefore, instantaneous speed is always equal to instantaneous velocity.

Look at the graphs (a) to (d) in below figure carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

(a) If a vertical line is drawn from any point on axis representing t, we see that it cuts the graph at two points which means particle has two positions at the same instant of time which is not possible. So, the given

(b) If a vertical line is drawn from any point on axis representing t, we see that it cuts the graph at two points which means particle has two velocities at the same instant of time which is not possible. So, the given

(c) Speed has negative values in the graph which is not possible. So, the given

(d) The given

Below figure shows the

No, because the x-t graph does not represent the trajectory of the path followed by a particle. From the graph, it is noted that at

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

Speed of the police van,

Muzzle speed of the bullet,

Speed of the thief’s car,

Since the bullet is fired from a moving van, its resultant speed can be obtained as:

= 150 + 8.33 = 158.33 m/s

Since both the vehicles are moving in the same direction, the velocity with which the bullet hits the thief’s car can be obtained as:

= 158.33 – 53.33 = 105 m/s

Suggest a suitable physical situation for each of the following graphs in below figure

(a) The given

(b) In the given

(c) The given

Below figure gives the

For simple harmonic motion (SHM) of a particle, acceleration (

Where ω = angular frequency

In this time interval,

In this time interval,

In this time interval,

Below figure gives the

The average speed of a particle shown in the

It is clear from the graph that the slope is maximum and minimum restively in intervals 3 and 2 respectively. Therefore, the average speed of the particle is the greatest in interval 3 and is the least in interval 2. The sign of average velocity is positive in both intervals 1 and 2 as the slope is positive in these intervals. However, it is negative in interval 3 because the slope is negative in this interval.

Figure 3.25 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 the points A, B, C and D?

i) As the change of speed is greatest in interval 2, So magnitude of acceleration is greatest in interval 2

ii) Height of the curve from the time-axis gives the average speed of the particle. It is apparent from the graph that height is the greatest in interval 3. Hence, average speed of the particle is the greatest in interval 3.

iii) v is positive in all the intervals

Acceleration is positive in interval 1 and 3 as speed is increasing while it is negative in interval 2 as speed is decreasing

iv) Points A, B, C, and D are all parallel to the time-axis. Hence, the slope is zero at these points. Therefore, at points A, B, C, and D, acceleration of the particle is zero.

A three-wheeler starts from rest, accelerates uniformly with 1 m/s

Distance covered by a body in

S

Where,

In the given case,

∴ S

Which is a linear equation, so graph would be a straight line

Now, substituting different values of

n |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |

S_{n} |
0.5 |
1.5 |
2.5 |
3.5 |
4.5 |
5.5 |
6.5 |
7.5 |
8.5 |
9.5 |

The plot between

After 10 secs, speed of the three-wheeler has become

V=10X1=10 m/s

So, it will cover 10 m in each sec which is represented by the horizontal line in the graph

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

Initial velocity of the ball,

Acceleration,

When the lift was stationary, the boy throws the ball.

Taking upward motion of the ball,

Final velocity,

From first equation of motion, time of ascent (

t = (

= -49

But, the time of ascent is equal to the time of descent.

Hence, the total time taken by the ball to return to the boy’s hand = 5 + 5 = 10 s.

The lift was moving up with a uniform velocity of 5 m/s. In this case, the relative velocity of the ball with respect to the boy remains the same i.e., 49 m/s. Therefore, in this case also, the ball will return back to the boy’s hand after 10 s.

On a long horizontally moving belt (below figure), a child runs to and fro with a speed 9 km /h (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h

(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 Solutions alter if motion is viewed by one of the parents?

(a)Speed of the belt,

Speed of the boy w.r.t belt,

Since the boy is running in the same direction of the motion of the belt, his speed (as observed by the stationary observer) can be obtained as:

(b)Since the boy is running in the direction opposite to the direction of the motion of the belt, his speed (as observed by the stationary observer) can be obtained as:

(c) Distance between the child’s parents = 50 m

As both parents are standing on the moving belt, the speed of the child in either direction as observed by the parents will remain the same i.e., 9 km/h = 2.5 m/s.

Hence, the time taken by the child to move towards one of his parents is given by

= 50 / 2.5 = 20 s

(d)If the motion is viewed by any one of the parents, Solutions obtained in (a) and (b) get altered. This is because the child and his parents are standing on the same belt and hence, are equally affected by the motion of the belt. Therefore, for both parents (irrespective of the direction of motion) the speed of the child remains the same i.e., 9 km/h.

For this reason, it can be concluded that the time taken by the child to reach any one of his parents remains unaltered.

Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m/s and 30 m/s. Verify that the graph shown in below figure 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 m s

Initial velocity,

Acceleration,

Using the relation,

Where, height of the cliff,

When this stone hits the ground,

– 5

t

Since the stone was projected at time

∴

Initial velocity,

Acceleration,

Using the relation,

x

= 200 + 30t - 5t

At the moment when this stone hits the ground;

– 5

Here again, the negative sign is meaningless.

∴

Subtracting equations (i) and (ii), we get

Equation (iii) represents the linear path of both stones. Due to this linear relation between (

Maximum separation between the two stones is at

(

This is in accordance with the given graph.

After 8 s, only second stone is in motion whose variation with time is given by the quadratic equation:

Hence, the equation of linear and curved path is given by

After 10 secs, both the stone are on the ground, so relative distance is zero and this is represented by horizontal line after 10 secs

The speed-time graph of a particle moving along a fixed direction is shown in below figure. Obtain the distance traversed by the particle between

(a)

(b)

(c)What is the average speed of the particle over the intervals in (a) and (b)?

(a) Distance travelled by the particle = Area under the given graph

= (1/2) × (10 - 0) × (12 - 0) = 60 m

Average speed = Distance

(b)Let

Total distance (

Let

Since the particle undergoes uniform acceleration in the interval

Where,

12 = 0 +a

Again, from first equation of motion, we have velocity at t=2sec

u

Distance travelled by the particle between time 2 s and 5 s i.e., in 3 s

= 4.8 × 3 + (1/2) × 2.4 × (3)

= 25.2 m ........(i)

Let

From first equation of motion,

0 = 12 +

Distance travelled by the particle in 1s (i.e., between

= 12 × 1 + (1/2) (-2.4) × (1)

= 12 - 1.2 = 10.8 m .........(ii)

From equations (i) and (ii) we get

Therefore,

Average speed = Total distance/time taken = 36 / 4 = 9 m/s

The velocity-time graph of a particle in one-dimensional motion is shown in below figure :

Which of the following formulae are correct for describing the motion of the particle over the time-interval

(a)

(b)

(c)

(d)

(e)

(f)

The correct formulae describing the motion of the particle are (c), (d) and, (f)

The given graph has a non-uniform slope. Hence, the formulae given in (a), (b), and (e) cannot describe the motion of the particle. Only relations given in (c), (d), and (f) are correct equations of motion.

Class 11 Maths Class 11 Physics Class 11 Chemistry