Class 11 Physics Chapter 2 Solutions | Motion in a Straight Line Explained

3.14

The distance from home to market = 2.5 km = 2500 m

The speed of the walking while going = 5 kmph = 1.388 m/s

The speed of walking back from market to home = 7.5 kmph = 2.083 m/s

(a) Magnitude of the average velocity = Displacement / time = 0, since the total displacement is zero.

(b)

i. Time taken to reach the market = distance / onward speed = 2500/1.388 = 1801 s = 30 minutes. So the average speed over 0 – 30 min is 5 kmph

ii. Time taken to reach back home = 2500 / 2.083 = 20 minutes. So the average speed = total distance covered / total time taken = (2500 +2500)/ (30 +20) m/min = 100 m/min = 6 kmph

iii. Average speed over the interval of time 0 to 40 mins = Distance covered / time taken = 2500 + (2.083 * 10 * 600) / (40 * 60) = 1.5624 m/s = 5.62 kmph

3.16

(a) Figure (a) shows two positions at the same time, which is not possible for a one-dimensional motion

(b) A particle cannot have velocity in two directions at the same time in one-dimensional motion

(c) The graph shows negative speed, which is not possible because speed is always positive

(d) Total path length decreases, which is not possible for a one – dimensional motion

3.18 Speed of the police van = 30 kmph = 8.33 m/s

Speed of the thief's car = 192 kmph = 53.33 m/s

The muzzle speed of the bullet = 150 m/s

Speed of the bullet = speed of the police van + muzzle speed of the bullet = 8.33 + 150m/s = 158.33 m/s

The relative velocity of the bullet = speed of the bullet – speed of the thief's car = 158.33 – 53.33 m/s = 105 m/s

3.6 The speed of the car = 126 kmph = (126 * 1000/3600) m/sec = 35 m/s

From the relation V² = U² + 2fs, where Final velocity V = 0, U = 35 m/s, s = 200m, we get retardation, f = -U²/2s= 3.06 m²/s

From the relation V = U – ft, we get t = U/f = 35/3.06 = 11.42 s.

3.21 The average speed in Interval 3 is the greatest and in Interval 2 is the least.

The average velocity is +ve in Interval 1 & 2 and –ve in Interval 3.

3.22 The change in the speed with time is maximum in interval 2. Therefore, the average acceleration is the greatest in magnitude in interval 2

The average speed is maximum in interval 3

The sign of velocity is positive in intervals 1,2 and 3

Acceleration depends on the slope. The acceleration is positive in interval 1 and 3, as the slope is positive. The acceleration is negative in interval 2, as the slope is negative

Acceleration at A, B, C and D is zero since the slope is parallel to the time axis at these instants

3.13

(a) Let us take the example of football world cup 2022. During Argentina match, Di Mario (A) passed the ball to Messi (B). Messi instantly passed the ball to Di Mario. Now the magnitude of displacement of the ball is 0 as the ball returns to Di Mario at the initial position. But the total length covered by the ball is AB + BA = 2AB. Hence total length covered by the particle (the ball) is more than the magnitude.

(b) If t is the time taken to cover the entire distance, the magnitude of the average velocity of the ball over time interval t = Magnitude of displacement / t = 0 / t = 0

The average speed of the ball = Total length of the path / time interval = 2AB / t

Thus the second quantity is greater than the first.

If the ball moves only in one direction from A to B, both the quantities are equal (considering one-dimensional motion)

3.10

(a) The direction of the acceleration is downward during the upward motion of the ball because the acceleration is due to gravity. Gravity always pulls the object toward the centre of the earth. 

(b) Velocity: At the peak the velocity is zero because the ball momentarily stops before falling downwards. 

(c) If we consider the highest point x = 0 m, t= 0 s,

During upward motion of the ball before it reaches the maximum height, x = +ve, velocity = -ve, acceleration = +ve. During downward motion, x = +ve, velocity = +ve, acceleration = +ve

(d) At the highest point, final velocity v = 0, initial velocity = 29.4 m/s, acceleration = 9.8m/s²

From the equation v2-u2 = 2gs, we get s = (29.4)²/2 * 9.8 = 44.1 m. From v –u = gt we get t = 29.4/9.8 s = 3 s. So the total time taken by the ball to reach player's hand = 3 * 2s = 6 s

(a) This graph is a Displacement-Time Graph (x- t) that shows the displacement increasing From A to B, and decreasing From B to C. The graph may represent a carom board where the striker hits the edge, rebounds with reduced speed, then moves in the opposite direction, hits the opposite wall and stops.

(b) This graph is a Velocity-Time Graph (v- t) that shows Sudden spikes and drops in velocity. There is a rapid change in velocity. This graph may represent a ball that falls on the ground from a certain height and rebounds with a reduced speed

(c) This graph is an Acceleration -Time Graph (a- t) that shows an instantaneous spike in acceleration. The graph looks like of a baseball, thrown on to a batter, moves with a uniform speed and then hit by the bat for a very short time interval.

3.17 For t<0, we cannot say that the particle moved in a straight line and for t>0 on a parabolic path as the x-t graph does not indicate. Instead, this x-t graph denotes a particle is dropped from a height at t=0

3.27

(a) Distance travelled by the particle between t = 0 s and t = 10 s is the area of the triangle = (1/2) x base x height = (1/2) x 10 x 12 = 60 m

The average speed of the particle is 60/10 m/s = 6 m/s

(b) Distance travelled by the particle between t = 2 s and t = 6 s

Let S₁ be the distance travelled by the particle in time 2 to 5 s and S₂ be the distance travelled between 5 to 6s.

For the motion  from 0 to 5 sec, u = 0, t = 5, v = 12 m/s

From the equation v = u + at we get a = (v-u)/t = 12/5 = 2.4 m/s²

Distance covered from 2 to 5 sec, S₁ = distance covered in 5 s – distance covered in 2 s

From the equation s = ut + at² we get,

= (1/2) * 2.4 * 25 – (1/2) * 2.4 * 4 = 30 – 4.8 = 25.2 m

For the motion from 5 sec to 10 sec, u = 12 m/s and a = -2.4 m/s²

and t = 5 sec to t = 6 sec means n = 1 for this motion

Distance covered in the 6th sec is S2 = u + (1/2) a (2n-1) = 12 + (1/2) (-2.4) (2 * 1-1) = 10.8 m

Therefore the total distance covered from t = 2 to 6 s = 25.2 + 10.8 m = 36 m

3.7 The uniform speed of A and B = 72 kmph = 20 m/s

The acceleration f = 1 m/s², time t = 50 s

Distance covered by train B = ut + 0.5 * ft²  = 20 * 50 + 0.5 * 1 * 50 ² = 2250 m

Distance covered by train A in the same period = ut = 50 * 20 = 1000 m

So the initial distance train A and train B was = 2250 – 1000 $–\left(400+400\right)$ = 450 m

3.11

(a) True – zero speed may have a non zero instantaneous non-zero acceleration

(b) False – zero speed will always make zero velocity

(c) True – constant speed will have zero acceleration

(d) False – If the positive value of acceleration is in the direction of its motion then it can speed up

3.15 Instantaneous speed and velocity are applicable for a small interval of time because the magnitude of the displacement is effectively equal to the distance travelled by the particle.

3.5 The speed of the jet plane, Vj = 500 kmph. If the speed of the product of combustion is Vpc, then V pc = Vpc-Vj = -1500 kmph (relative to the jet plane)

Speed of the combustion product relative to the observer on the ground,

Vpc = -1500 + Vj = -1500+500 = -1000 kmph

3.20 In simple harmonic motion, the acceleration is expressed as a = - $?$ 2x, where $?$ is the angular frequency.

(a) At t = 0.3 s, position x is negative, velocity v is negative and acceleration a ( from above equation) will be positive

(b) At t = 1.2 s, position x is positive, velocity is positive, acceleration a will be negative

(c) At t = -1.2 s, position x is negative, velocity is positive and acceleration will be positive.

3.28 The graph has a non-uniform slope between the intervals t₁ and t₂ – the graph is not a straight line. The equations (a), (b), and (e) do not describe the motion of the particle. Only the relations (c), (d) and (f) are correct.   

3.23 The distance covered by the 3 wheeler on a straight line in the nth second can be expressed as:

Sn = u + a (2n-1)/2 …… (1),

Where

a = acceleration

u = initial velocity

n = time = 1,2,3, ……., n

Given, u = 0, a = 1m/s², from equation (1) we get Sn = (2n-1)/2 ……. (2)

With various values of n, we get Sn

            n                      Sn

            1                      0.5

            2                      1.5

            3                      2.5

            4                      3.5

            5                      4.5

            6                      5.5

            7                      6.5

            8                      7.5

            9                      8.5

            10                    9.5

3.4

The time taken for each step is 1s, he covers 5 m in 5 secs and goes backward by 3 m in 3 secs.

So in 8 s he covers 2 m. To cover 13m distance he needs to complete 8 m (13-5). To cover 8m, he will require 8 * 4 = 32 s. To fall in a pit, he needs to move another 5m in 5s.

So total time taken = 32 +5 = 37 s

Let’s have a look at how we came to this answer. 

We have to calculate how long it will take for the drunkard to fall into a pit located 13 metres from his starting point.

In each cycle of movement, the drunkard takes 5 steps forward and 3 steps backward. We calculate the net displacement per cycle. 

Time for one cycle = 5 seconds (forward) + 3 seconds (backward) = 8 seconds.

Net displacement in one cycle = 5 meters (forward) - 3 meters (backward) = 2 meters.

Reaching the pit: The pit is 13 metres away. We can determine the time taken by analysing his position after each cycle.

After 1 cycle (8 s): Position = 2 m

After 2 cycles (16 s): Position = 4 m

After 3 cycles (24 s): Position = 6 m

After 4 cycles (32 s): Position = 8 m

At the 32-second mark, the drunkard is at the 8-metre position. This we get to know when learning about instantaneous velocity. 

In his next set of forward steps, he will cover 5 metres in 5 seconds. This means he will reach the 13-metre mark and fall into the pit. 

Total time calculation: The total time is the time taken to reach the 8-metre mark plus the time to take the final 5 steps.

Total time = 32 seconds + 5 seconds = 37 seconds.

Practice more NCERT Solutions for Motion in a Straight Line chapter.

3.9 Cycling speed $V_c$ = 20 kmph

Let as assume speed of the bus is $V_b$

The relative velocity of the buses in the direction of motion of the cyclist = $V_b$ - $V_c$

Buses go past him every 18 mins = 18/60h = 0.3h

Distance covered = ( $V_b-V_c$ ) × 0.3

Since the buses leave every T minutes, ( $V_b-V_c$ ) × 0.3 = $V_b$ × (T/60) ……(1)

The relative velocity of the buses in the opposite direction of the cyclist = $V_b$ + $V_c$

Buses go past him in the opposite direction = every 6 m = 6/60 h

Distance covered = ( $V_b$ + $V_c$ ) × 1/10

Since the buses leave every T minute, we get ( $V_b$ + $V_c$ ) × 1/10 = $V_b$ × T/60 …….(2)

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

( $V_b$ + $V_c$ ) / 10 = ( $V_b-V_c$ ) × 0.3

$V_b$ + $V_c$ = 3( $V_b$ - 3 $V_c$ )

2 $V_b$ = 4 $V_c$ , $V_b$ = 2 $V_c$ So $V_b$ = 2 × 20 = 40 kmph

From the equation (1), we get

(40-20) × 0.3 = 40 × T/60, T = 9 min

This question belongs to the Motion in a straight line chapter that we will be solving this with explanation.

(a) A railway carriage moving without jerks between two stations

Whenever a railway carriage moves between two stations smoothly without jerks, it is considered as a point object. This is due to the fact that distance between the two stations is larger as compared to the size of carriage. However, we are interested in the overall motion instead of the movement of specific parts of carriage.

(b) A monkey sitting on top of a man cycling smoothly on a circular track

Here, both monkey as well as the cyclist are considered as a single point object if we analyze their motion on the circular track. The radius of the circular path is much larger as compared to the size of the monkey and cyclist. This makes the point object approximation valid for studying overall motion.

There are more complex and other important questions in the NCERT excercise of Motion in a Straight Line that you can check out for verifying if you are answering questions correctly. 

(c) A spinning cricket ball that turns sharply on hitting the ground

A spinning cricket ball that turns sharply on hitting the ground is not a point object. The spin and sharp turn are highly influenced by the size, shape and rotation of the ball. This affects the trajectory of the ball and its interaction with the ground.

(d) A tumbling beaker that has slipped off the edge of a table

A tumbling beaker which has slipped off a table is not a point object since it involves rotation and changes in the orientation. These depend on the shape and size of beaker. 

Students must go through those solutions if they are currently studying in class 12th and soon they have to take the 12th CBSE board exam. 

3.8

Speed of car A, Va = 36 kmph = 10m/s

Speed of car B, Vb = 54 kmph = 15 m/s

Speed of car C, Vc = -54 kmph = -15 m/s

Relative speed of A, w.r.t. C = Va – (-Vb) = 10 +15 = 25 m/s

Relative speed of B, w.r.t. A = Vb-Va = 15-10 = 5 m/s

Distance between AB & AC = 1 km = 1000m

Time taken by the car C to cover the distance AC, from s = ut

We get t = 1000/25 = 40s

To avoid collision with C, the minimum acceleration the vehicle B must have, from the relation, s = ut + (1/2) ft²

We get 1000 = 5 * 40 + (1/2) f * 40² f = 1 m/s²

3.3 Office distance = 2.5 km

Walking speed = 5 kph

Time taken to reach office = 5/2.5 = 0.5 h

OA is the path to reach office

AB is the duration of office stay

Auto speed = 25 kph

Time taken by auto = 2.5/25 h = 6 minutes

BC is the path for returning

On the position-time graph, this is a straight line sloping downwards. It starts at the point representing 5:00 PM and 2.5 km, and ends at 5:06 PM at 0 km. It tells us that the woman has arrived home. 

The slope of this line represents instantaneous velocity. Because the return trip is by the auto at a much higher speed (25 km/h), the slope of line BC is significantly steeper than the line for the outward journey. A steeper slope on a position-time graph denotes a higher speed.

Master such questions with our NCERT Solutions for Motion in a Straight Line. 

3.2 

(a) A lives closer to the school than B ( OP

(b) A starts earlier than B from school. From the graph, A starts from O at time 0 whereas B starts after some finite value t

(c) B walks faster than as the slope for B is steeper than A

(d) Both B and A reaches home at the same time

(e) At the intersection of the two lines, it is evident that B overtakes A once

3.24 The initial velocity of the ball, u = 49m/s

First Case: When the ball returns to his hands, total displacement = 0

From the relation s = ut + 0.5at², we get 0 = 49t + 0.5 (-9.81) t²

4.905 = 49t      Hence t = 10s

Second Case:

As the lift started moving up with a speed of 5 m/s, the initial velocity of the ball = 49 + 5 m/s = 54 m/s

If t' is time for the ball to return to his hand, the displacement of the ball will be = 5t'

From the relation        s = ut + 0.5 x at², we get

5t' = 54t' + 0.5 * (-9.8) t'²

49t' = 4.9 t'²

t' = 10 s      

Initial velocity, u₁ = 15m/s, acceleration, a = -g = -10 m/s

From the relation  s₁=s₀+u₁t+ (1/2)at² where

s₀ = cliff height, s₁ = total height of the fall of the first stone, we get

s₁ = 200 + 15t – 5t²  ………. (1)

When the stone hit the floor, s₁ = 0, so the equation (1) becomes

0 = 200 +15t - 5t² = t² -3t – 40 = (t-8) (t+5) = 0

So t = 8s or -5s

Since the stone was thrown at t=0, so t cannot be –ve. Hence t = 8s

For the second stone,

Initial velocity, u₁ = 30 m/s, acceleration, a = -g = -10 m/s

From the relation  s₂=s₀+u₁t+ (1/2)at² where

s₀ = cliff height, s₂= total height of the fall of the second stone, we get

s₂ = 200 + 30t – 5t²  ………. (2)

When the stone hit the floor, s₂ = 0, so the equation (2) becomes

0 = 200 +30t - 5t² = t² -6t – 40 = (t-10) (t+4) = 0

So t = 10s or -4s

Since the stone was thrown at t=0, so t cannot be –ve. Hence t = 10s

Subtracting equation (1) from equation (2), we get

s_{2 –} s₁ = (200 + 30t – 5t²  ) – (200 + 15t – 5t² )  = 15t ……. (3)

Equation (3) represents the linear trajectory of the two stones because to this linear relation between (s_{2 –} s₁₎ and t, the projection is straight line till 8s (from the graph)

The maximum distance between the two stones is at t = 8s. So

(s_{2 –} s₁) _max = 15 x 8 = 120 m. This value has been depicted correctly in the graph.

After 8s, only the second stone is in motion (t for first stone has been found out to be 8s), the trajectory of second stone is obtained in equation (2), = 200 + 30t – 5t²  , which is a curved path.