Important Questions for CBSE Class 11 Physics Chapter 3 - Motion in a Straight Line

1 Mark Questions

1. Under what condition is the relation  $ s=vt $  correct?

Ans: The relation  $ s=vt $  is valid when the particle moves with uniform velocity and along a straight line.

2. Two balls of different masses are thrown vertically upward with the same initial speed. Which one will rise to a greater height?

Ans: When two balls of different masses are thrown vertically upward with the same initial speed, both of them will rise to a greater height.

3. What is the relative velocity of two bodies having equal velocities?

Ans: The relative velocity of two bodies having equal velocities is given by:

Consider, ${{v}_{a}}={{v}_{b}}=v$

Then, ${{v}_{ab}}={{v}_{a}}-{{v}_{b}}=v-v=0$

4. A 400m long railway train, 400m long is going from New Delhi railway station to Kanpur. Can we consider a railway train as a point object?

Ans: Yes, we consider a railway train as a point object. This is because the length of the train is smaller as compared to the distance between New Delhi and Kanpur.

5. Shipra went from her home to school 2.5km away. On finding her home closed she returned to her home immediately. What is her net displacement? What is the total distance covered by her?

Ans: From the above question it can be interpreted that:

$ Displacement\text{ }=\text{ }0km $  And

$ Distance\text{ }=\text{ }2.5km\text{ }+\text{ }2.5km\text{ }=\text{ }5.0km $.

6. Can the speed of an object be negative? Justify.

Ans: The speed of an object can never be negative. This is because the distance is also always positive.

7. Under what condition the displacement and the distance of a moving object will have the same magnitude?

Ans: Distance and displacement have the same magnitude when the object will move in a straight line.

8. What is the shape of the displacement time graph for uniform linear motion?

Ans: The shape of the displacement time graph for uniform linear motion is a straight line inclined to the time axis (x-axis).

9. Figure shows a displacement time graph. Comment on the sign of velocities at the point

P, Q, R, S, and T.

(Image Will Be Updated Soon)

Ans: From the figure,

Velocity at P and T are positive.

Velocity at Q and S is zero.

Velocity at R is negative.

10. The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29. Which of the following formulae are correct for describing the motion of the particle over the time-interval  $ {{t}_{1}}\text{ }to\text{ }{{t}_{2}} $:

(Image Will Be Updated Soon)

a) $x\left( {{t}_{2}} \right)=x\left( {{t}_{1}} \right)+v\left( {{t}_{1}} \right)\left( {{t}_{2}}-{{t}_{1}} \right)+\left( 1/2 \right)a{{\left( {{t}_{2}}-{{t}_{1}} \right)}^{2}}$

b) $v\left( {{t}_{2}} \right)=v\left( {{t}_{1}} \right)+a\left( {{t}_{2}}-{{t}_{1}} \right)$

c) ${{v}_{average}}=\left( x\left( {{t}_{2}} \right)-x\left( {{t}_{1}} \right) \right)/\left( {{t}_{2}}-{{t}_{1}} \right)$

d) ${{a}_{average}}=\left( v\left( {{t}_{2}} \right)-v\left( {{t}_{1}} \right) \right)/\left( {{t}_{2}}-{{t}_{1}} \right)$

e) $x\left( {{t}_{2}} \right)=x\left( {{t}_{1}} \right)+{{v}_{average}}\left( {{t}_{2}}-{{t}_{1}} \right)+\left( 1/2 \right){{a}_{average}}{{\left( {{t}_{2}}-{{t}_{1}} \right)}^{2}}$

$x\left( {{t}_{2}} \right)-x\left( {{t}_{1}} \right)=\text{area under the v-t curve bounded by the t-axis }$ 

$\text{and the dotted line shown}\text{.}$

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

11. 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.

Ans: As the size of a carriage is very small as compared to the distance between two stations, the carriage can be treated as a point-sized object.

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

Ans: As the size of a monkey is very small as compared to the size of a circular track, the monkey can be considered as a point-sized object on the track.

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

Ans: As the size of a spinning cricket ball is comparable to the distance through which it turns sharply on hitting the ground, the cricket ball cannot be considered as a point object.

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

Ans: As the size of a beaker is comparable to the height of the table from which it slipped, the beaker cannot be considered as a point object.

2 Marks Questions

1. Write the characteristics of displacement.

Ans: Following are the characteristics of displacement:

(1) Displacement is a vector quantity having both magnitude and direction.

(2) Displacement of a given body can be positive, negative, or zero.

2. Draw a displacement time graph for uniformly accelerated motion. What is its shape?

Ans: Following is the time graph for uniformly accelerated motion which is parabolic in shape.

(Image Will Be Updated Soon)

3. Sameer went on his bike from Delhi to Gurgaon at a speed of 60km/hr and came back at a speed of 40km/hr. What is his average speed for the entire journey?

Ans: In the above question it is given that:

Speed of the bike when Sameer traveled from Delhi to Gurgaon is ${{v}_{1}}=60km/hr$.

Come back speed is ${{v}_{2}}=40km/hr$.

Therefore, average speed will be:

$ Average\text{ }speed=\dfrac{2{{v}_{1}}{{v}_{2}}}{{{v}_{1}}+{{v}_{2}}}=\dfrac{2\left( 60 \right)\left( 40 \right)}{60+40}=48km/hr $

4. What causes variation in the velocity of a particle?

Ans: Variation in velocity of a particle happens when:

(1) magnitude of velocity changes

(2) direction of motion changes.

5. Figure. Shows displacement – time curves I and II. What conclusions do you draw from these graphs?

(Image Will Be Updated Soon)

Ans: From the graph given in the question we can conclude that: 

(1) Both the curves represent uniform linear motion.

(2) Uniform velocity of II is more than the velocity of I because the slope of the curve (II) is greater.

6. Displacement of a particle is given by the expression  $ x=3{{t}^{2}}+7t9 $, where x is in meters and t is in seconds. What is acceleration?

Ans: Expression of Displacement a particle is given by  $ x=3{{t}^{2}}+7t9 $ 

Therefore,

$v=\dfrac{dx}{dt}=6t+7$

And

$a=\dfrac{dv}{dt}=6m/{{s}^{2}}$, which is the required acceleration.

7. A particle is thrown upwards. It attains a height (h) after 5 seconds and again after 9s comes back. What is the speed of the particle at a height h?

Ans: According to Newton’s laws of motion:

$s=ut+\dfrac{1}{2}a{{t}^{2}}$

The net displacement at  $ 4s $  is zero as the particle comes to the same point at  $ 9s $  where it was at  $ 5s $ .

Hence,

$0=\left( u\times 4 \right)-\dfrac{1}{2}\left( g \right){{\left( 4 \right)}^{2}}$

$\left( u \right)\times 4=\dfrac{1}{2}\left( g \right){{\left( 4 \right)}^{2}}$

Hence,

$u=2\times 9.8=19.6m/s$

Hence, the speed of the particle at a height h is $19.6m/s$.

8. Draw displacement time graph for a uniformly accelerated motion? What is its shape?

Ans: Following is the displacement time graph for a uniformly accelerated motion:

(Image Will Be Updated Soon)

9. The displacement x of a particle moving in one dimension under the action of the constant force is related to the time by the equation $t=\sqrt{x}-3$ where x is in meters and t is in seconds. Find the velocity of the particle at (1) t = 3s (2) t = 6s.

Ans: The given equation for displacement of particle is

$t=\sqrt{x}-3$

$\Rightarrow x={{\left( t+3 \right)}^{2}}$

Now, velocity is calculated as:

$v=\dfrac{dx}{dt}=2\left( t+3 \right)$

Therefore,

i) v at $t=3s$ will be:

$v=2\left( 3+3 \right)=12m/s$

And

ii) v at $t=6s$ will be:

$v=2\left( 6+3 \right)=18m/s$

10. A balloon is ascending at the rate of 4.9m/s. A pocket is dropped from the balloon when situated at a height of 245m. How long does it take the packet to reach the ground? What is its final velocity?

Ans: In the above question it is given that:

Initial velocity, $u=4.9m/s$

Height, $h=245m$.

As the pocket is in freefall, $a=g=9.8m/{{s}^{2}}$.

Therefore, using Newton’s Laws of motion:

$s=ut+\dfrac{1}{2}a{{t}^{2}}$

$245=-4.9t+\dfrac{1}{2}\left( 9.8 \right){{t}^{2}}$

$4.9{{t}^{2}}-4.9t-245=0$

$\Rightarrow t=7.\text{6 or }-5.6$

Hence, $t=7.6s$

Now, $v=u+at$

$v=-4.9+9.8\left( 7.6 \right)=69.6m/s$

Hence, it takes $7.6s$ for the packet to reach the ground and the final velocity is $69.6m/s$.

11. A car moving on a straight highway with a speed of 126km/hr. is brought to a stop within a distance of 200m. What is the retardation of the car (assumed uniform) and how long does it take for the car to stop?

Ans: In the above question it is given that:

$u=126km/r=35m/s$,

$v=0m/s$,

$s=200m$And

$t=?$

Now,

We know that:

$a=\dfrac{{{u}^{2}}-{{v}^{2}}}{2s}=\dfrac{{{35}^{2}}-0}{2\left( 200 \right)}=-3.06m/{{s}^{2}}$

Now, $v=u+at$

Therefore, $0=35-\left( 3.06 \right)t$

$t=11.4s$

Hence, the retardation of the car is $-3.06m/{{s}^{2}}$ and it takes $11.4s$ for the car to stop.

12. 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?

Ans: We know that instantaneous velocity is the first derivative of distance with respect to time.

Here, the time interval is so small that it is assumed that the particle does not change its direction of motion. Therefore, both the total path length and magnitude of displacement become equal in this interval of time. Thus, instantaneous speed is always equal to instantaneous velocity.

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

(Image Will Be Updated Soon)

Ans:

a) Consider the x-t graph, given in fig (a). It does not represent the one-dimensional motion of the particle. This is because a particle cannot have two positions at the same instant of time.

b) Consider the x-t graph, given in fig (b). It does not represent the one-dimensional motion of the particle. This is because a particle can never have two values of velocity at the same instant of time.

c) Consider the x-t graph, given in fig (c). It does not represent the one-dimensional motion of the particle. This is because speed being a scalar quantity cannot be negative.

d) Consider the x-t graph, given in fig (d). It does not represent the one-dimensional motion of the particle. This is because the total path length traveled by the particle cannot decrease with time.

14. Figure shows the x-t plot of the 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.

(Image Will Be Updated Soon)

Ans: No, this is because the x-t graph does not represent the trajectory of the path followed by a particle. Also from the graph, it is clear that at  $ t=0,\text{ }x=0 $.

15. A police van moving on a highway with a speed of  $ 30\text{ }km/hr $  fires a bullet at a thief’s car speeding away in the same direction with a speed of  $ 192\text{ }km/hr $ . If the muzzle speed of the bullet is  $ 150\text{ }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).

Ans: In the above question it is given that:

Speed of the police van is  $ {{v}_{p}}=30\text{ }km/hr=8.33m/s $ .

Muzzle speed of the bullet is  $ {{v}_{b}}=150\text{ }m/s $ .

Speed of the thief’s car is  $ {{v}_{t}}=192\text{ }km/hr=53.33m/s $ .

As the bullet is fired from a moving van, its resultant speed will be $150+8.33=158.33m/s$.

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:

${{v}_{bt}}={{v}_{b}}-{{v}_{t}}=158.33-53.33=105m/s$

16. 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.

(Image Will Be Updated Soon)

Ans: The average speed is greatest in interval 3 and least in interval 2. It is positive in intervals 1 & 2 and negative in interval 3.

The average speed of a particle shown in the x-t graph is given by the slope of the graph in a particular interval of time.

From the graph, it is clear that the slope is maximum and minimum restively in intervals 3 and 2 respectively. Thus, 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.

17. A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to  $ 49\text{ }m/s $ . How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of $5m/s$ 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?

Ans: In the above question it is given that:

Initial velocity of the ball is  $ u\text{ }=\text{ }49\text{ }m/s $ .

Acceleration is $a=-g=-9.8m/{{s}^{2}}$.

Consider
case I:

When the lift was stationary, the boy threw the ball. Taking upward motion of the ball,

Final velocity, v of the ball becomes zero at the highest point.

From first equation of motion, time of ascent (t) is given as:

$v=u+at$

$t=\dfrac{v-u}{a}=\dfrac{-49}{-9.8}=5s$

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

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

Case II:

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

18. A jet airplane travelling at the speed of  $ 500\text{ }km/hr $  ejects its products of combustion at the speed of  $ 1500\text{ }km/hr $  relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?

Ans: In the above question it is given that:

Speed of the jet airplane,  $ =\text{ }500\text{ }km/hr $ 

Relative speed of its products of combustion with respect to the plane, $ =1500\text{ }km/hr $ .

Speed of its products of combustion with respect to the ground $=1500-500=1000km/hr$

Hence, the speed of the latter with respect to an observer on the ground is $1000km/hr$.

19. Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of  $ 72\text{ }km/hr $  in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by  $ 1\text{ }m\text{ }{{s}^{-2}} $ . If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?

Ans: In the above question it is given that:

For train A: Initial velocity is  $ u\text{ }=\text{ }72\text{ }km/h\text{ }=\text{ }20\text{ }m/s $ 

Time is  $ t=50\text{ }s $ .

Acceleration, ${{a}_{1}}=0$ (Since it is moving with a uniform velocity)

From second equation of motion, distance $\left( {{s}_{1}} \right)$ covered by train A can be obtained as:

$s=ut+\left( 1/2 \right)a{{t}^{2}}$

$=20\times 50+0=1000m$

For train B:

Initial velocity is  $ u\text{ }=\text{ }72\text{ }km/h\text{ }=\text{ }20\text{ }m/s $ 

Acceleration, $a=1m/{{s}^{2}}$

Time is  $ t=50\text{ }s $ .

From second equation of motion, distance $\left( {{s}_{n}} \right)$ covered by train A can be obtained as:

${{s}_{n}}=ut+\left( 1/2 \right)a{{t}^{2}}$

$=20\times 50+\left( 1/2 \right)\left( 1 \right){{\left( 50 \right)}^{2}}=2250m$

Hence,

$ Length\text{ }of\text{ }both\text{ }trains\text{ }=\text{ }2\text{ }\times \text{ }400\text{ }m\text{ }=\text{ }800\text{ }m $ 

Therefore, the original distance between the driver of train A and the guard of train B is  $ 2250-1000\text{ }-\text{ }800\text{ }=\text{ }450m. $

3 Marks Questions

1. Define, $v=u+at$ from the velocity-time graph.

Ans: Consider the graph given below

(Image Will Be Updated Soon)

Slope of graph is

$\tan \theta =\dfrac{u-v}{t}$

And $\tan \theta =a$

Hence, $at=u-v$

$v=u+at$

2. A particle is moving along a straight line and its position is given by the relation

$x=\left( {{t}^{3}}-6{{t}^{2}}-15t+40 \right)m$

Find 

a) The time at which velocity is zero.

b) Position and displacement of the particle at that point.

c) Acceleration

Ans: Given expression of position is $x=\left( {{t}^{3}}-6{{t}^{2}}-15t+40 \right)m$.

$v=\dfrac{dx}{dt}=\left( 3{{t}^{2}}-12t-15 \right)m/s$ and

$a=\dfrac{dv}{dt}=\left( 6t-12 \right)m/{{s}^{2}}$

a) Calculating time at which velocity is zero,

$\left( 3{{t}^{2}}-12t-15 \right)=0$

${{t}^{2}}-4t-5=0$

$\left( t-5 \right)\left( t+1 \right)=0$

$\therefore t=5,-1$

Hence, $t=5s$

b) Position at $t=5s$ is given by

$x={{\left( 5 \right)}^{3}}-6{{\left( 5 \right)}^{2}}-15\left( 5 \right)+40=-60$

Position at $t=0s$ is given by

$x={{\left( 0 \right)}^{3}}-6{{\left( 0 \right)}^{2}}-15\left( 0 \right)+40=40$

Hence displacement$={{x}_{5}}-{{x}_{0}}=-60-40=-100m$

c) Acceleration at $t=5s$ is given by:

$a=6\left( 5 \right)-12=18m/{{s}^{2}}$

3. A police jeep on a petrol duty on national highway was moving with a speed of 54km/hr. in the same direction. It finds a thief rushing up in a car at a rate of 126km/hr in the same direction. Police sub-inspector fired at the car of the thief with his service revolver with a muzzle speed of 100m/s. With what speed will the bullet hit the thief's car?

Ans: In the above question it is given that:

${{V}_{PJ}}=54km/hr=15m/s$

${{V}_{TC}}=126km/hr=35m/s$

${{v}_{b}}=100m/s$

Hence,

Velocity of car w.r.t. police, ${{V}_{CP}}=35-15=20m/s$

Velocity of bullet w.r.t. car, ${{V}_{BC}}=100-20=80m/s$

Hence, the bullet will hit the car with velocity $80m/s$.

4. Establish the relation ${{S}_{nth}}=u+\dfrac{a}{2}\left( 2n-1 \right)$ where the letters have their usual meanings.

Ans: We have, ${{S}_{nth}}=u+\dfrac{a}{2}\left( 2n-1 \right)$.

We know that ${{S}_{nth}}={{S}_{n}}-{{S}_{n-1}}$

And

${{S}_{n}}=un+\dfrac{1}{2}a{{n}^{2}}$

${{S}_{n-1}}=u\left( n-1 \right)+\dfrac{1}{2}a{{\left( n-1 \right)}^{2}}$

$ {{S}_{nth}}=un+\dfrac{1}{2}a{{n}^{2}}-u\left( n-1 \right)-\dfrac{1}{2}a{{\left( n-1 \right)}^{2}} $ 

Hence,

$ {{S}_{nth}}=u-\dfrac{1}{2}a+na $ 

Therefore, 

$ {{S}_{nth}}=u+\dfrac{a}{2}\left( 2n-1 \right) $ 

Hence proved.

5. A stone is dropped from the top of a cliff and is found to travel 44.1m diving at the last second before it reaches the ground. What is the height of the cliff? g = 9.8m/s2

Ans: Consider the height of the cliff to be h m.

$u=0m/s$,

$a=g=9.8m/{{s}^{2}}$.

If n is the total time taken by the stone while falling,

$ {{S}_{nth}}=u+\dfrac{a}{2}\left( 2n-1 \right) $ 

$ 44.1=0+\dfrac{9.8}{2}\left( 2n-1 \right) $ 

$n=\dfrac{10}{2}=5s$

Now,

$h=ut+\dfrac{1}{2}a{{t}^{2}}$ 

$h=\dfrac{1}{2}\left( 9.8 \right){{\left( 5 \right)}^{2}}=122.5m$, which is the required height.

6. Establish $S=ut+\dfrac{1}{2}a{{t}^{2}}$ from the velocity time graph, a uniform accelerated motion.

(Image Will Be Updated Soon)

Ans: Consider the graph given in the question.

The displacement of the particle is given by the area under the v-t graph.

$ S\text{ }=\text{ }area\text{ }OABC $ 

$S\text{ }=\text{ }area\text{ }of\text{ }rectangle\text{ }AODC\text{ }+\text{ }area\text{ }of\text{ }ADB$

Hence,

$S=\left( OA\times OC \right)+\left( \dfrac{1}{2}AD\times BD \right)$

$S=ut+\dfrac{1}{2}\left( AD \right)\times \left( \dfrac{AD\times DB}{AD} \right)$

$S=ut+\dfrac{1}{2}{{\left( AD \right)}^{2}}\times \left( \dfrac{DB}{AD} \right)$

$S=ut+\dfrac{1}{2}{{\left( t \right)}^{2}}\times \left( \dfrac{DB}{AD} \right)$

$S=ut+\dfrac{1}{2}{{\left( t \right)}^{2}}\times \left( a \right)$

$\left[ \text{As }a=\tan \theta =\dfrac{BD}{AD} \right]$

Therefore,

$S=ut+\dfrac{1}{2}a{{t}^{2}}$

7. 

a) Define the term relative velocity.

Ans: The relative velocity of any object A with respect to object B is termed as the time rate of change of position of A with respect to B.

b) Write the expression for the relative velocity of one moving with respect to another body when objects are moving in the same direction and are moving in opposite directions?

Ans: Consider two objects to be moving in the same direction,

Then,

${{\overset{\to }{\mathop{V}}\,}_{AB}}={{\overset{\to }{\mathop{V}}\,}_{A}}-{{\overset{\to }{\mathop{V}}\,}_{B}}$

Where,

${{\overset{\to }{\mathop{V}}\,}_{A}}$ is the velocity of A.

${{\overset{\to }{\mathop{V}}\,}_{B}}$ is the velocity of B.

${{\overset{\to }{\mathop{V}}\,}_{AB}}$ is the velocity of A with respect to B.

Now consider two objects to be moving in the opposite direction,

Then,

$ {{\overset{\to }{\mathop{V}}\,}_{AB}}={{\overset{\to }{\mathop{V}}\,}_{A}}-\left( {{\overset{\to }{\mathop{-V}}\,}_{B}} \right)=\overset{\to }{\mathop{{{V}_{A}}}}\,+\overset{\to }{\mathop{{{V}_{B}}}}\, $ 

Where,  $ -\overset{\to }{\mathop{{{V}_{B}}}}\, $  indicates that B moves in the direction opposite to A.

b) A Jet airplane traveling at the speed of 500km/hr ejects its products of combustion at the speed of 1500km/h relative to the Jetplane. What is the speed of the latter with respect to an observer on the ground?

Ans: We have,

Velocity of the Jet plane,  $ {{V}_{J}}=\text{ }500km/hr $ 

Velocity of gases w.r.t. Jet plane  $ {{V}_{gJ}}\text{ }=\text{ }-1500km/hr $  (direction is opposite) $ {{\overset{\to }{\mathop{V}}\,}_{gJ}}={{\overset{\to }{\mathop{V}}\,}_{g}}-\left( {{\overset{\to }{\mathop{-V}}\,}_{J}} \right)=\overset{\to }{\mathop{{{V}_{g}}}}\,+\overset{\to }{\mathop{{{V}_{J}}}}\, $ 

Now, we know that hot gases also come out in opposite direction of the Jet plane,

Velocity of the gas, ${{V}_{g}}=-1500+500=-1000km/hr$.

8. Define (i)  $ v\text{ }=\text{ }u\text{ }+\text{ }at $  (ii)  $ {{v}^{2}}\text{ }\text{ }{{u}^{2}}=\text{ }2as $  by calculus method.

Ans:

i) Acceleration is given by:

$a=\dfrac{dv}{dt}$

$\Rightarrow dv=adt$

Integrating on both sides,

$\int{dv}=\int{adt}$

Hence, $v=at+k$ …… (1)

Where, k is the constant.

When, $t=0$ and $\theta =u$

We get $k=u$

Hence,  $ v\text{ }=\text{ }u\text{ }+\text{ }at $ .

ii) We have, $a=\dfrac{dv}{dt}$

Multiplying and dividing by dx,

$a=\dfrac{dv}{dt}\times \dfrac{dx}{dx}$

$a=\dfrac{dv}{dt}\times \theta $

$adx=vdv$

As $\dfrac{dx}{dt}=v$

On integrating,

$a\int\limits_{{{x}_{0}}}^{x}{dx}=\int\limits_{v}^{\theta }{vdv}$

Hence, $a\left( x-{{x}_{0}} \right)=\dfrac{{{u}^{2}}}{2}-\dfrac{{{v}^{2}}}{2}$ 

As $\left( x-{{x}_{0}}=s \right)$

$as=\dfrac{{{u}^{2}}-{{v}^{2}}}{2}$

 $ {{v}^{2}}\text{ } - \text{ }{{u}^{2}}=\text{ }2as $

9. A woman starts from her home at 9.00 am, walks with a speed of  $ 5\text{ }km/hr $  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\text{ }km/hr $. Choose suitable scales and plot the x-t graph of her motion

Ans: In the above question it is given that:

Speed of the woman  $ =\text{ }5\text{ }km/h $ .

Distance between her office and home  $ =\text{ }2.5\text{ }km $ .

$ Time\text{ }taken\text{ }=\text{ }Distance\text{ }/\text{ }Speed $ 

$ =\text{ }2.5\text{ }/\text{ }5\text{ }=\text{ }0.5\text{ }h\text{ }=\text{ }30\text{ }min $ 

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

Now,  $ speed\text{ }of\text{ }the\text{ }auto\text{ }=\text{ }25\text{ }km/h $ .

$ Time\text{ }taken\text{ }=\text{ }Distance\text{ }/\text{ }Speed $ 

$ =\text{ }2.5\text{ }/\text{ }25\text{ }=\text{ }1\text{ }/\text{ }10\text{ }=\text{ }0.1\text{ }h\text{ }=\text{ }6\text{ }min $ 

The suitable x-t graph of the motion of the woman is shown in the given figure.

(Image Will Be Updated Soon)

10. 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.

Ans: In the above question it is given that:

Distance covered with 1 step  $ =\text{ }1\text{ }m $ 

Time taken  $ =\text{ }1\text{ }s $ 

Time taken to move first  $ 5\text{ }m $  forward  $ =\text{ }5\text{ }s $ 

Time taken to move  $ 3\text{ }m $ backward  $ =\text{ }3\text{ }s $ 

$ Net\text{ }distance\text{ }covered\text{ }=\text{ }5\text{ }\text{ }3\text{ }=\text{ }2\text{ }m $ 

Net time taken to cover  $ 2\text{ }m\text{ }=\text{ }8\text{ }s $ .

Hence,

Drunkard covers  $ 2\text{ }m $ in  $ 8\text{ }s $ .

Drunkard covered  $ 4\text{ }m $ in  $ 16\text{ }s $ .

Drunkard covered  $ 6\text{ }m $ in $2\text{4 s}$.

Drunkard covered  $ 8\text{ }m $ in  $ 32\text{ }s $ .

In the next  $ 5\text{ }s $ , the drunkard will cover a distance of  $ 5\text{ }m $  and a total distance of  $ 13\text{ }m $  and then fall into the pit.

Net time taken by the drunkard to cover $km$ $ 13\text{ }m\text{ }=\text{ }32\text{ }+\text{ }5\text{ }=\text{ }37\text{ }s $ 

The x-t graph of the drunkard’s motion can be shown below:

(Image Will Be Updated Soon)

11. A car moving along a straight highway with a speed of  $ 126\text{ }km/hr $  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?

Ans: In the above question it is given that:

Initial velocity of the car is  $ u=126\text{ }km/hr=35m/s $ . 

Final velocity of the car is $v=0km/hr$.

Distance covered by the car before coming to rest is  $ 200\text{ }m $ .

Consider retardation produced in the car $=a$

From third equation of motion, ${{v}^{2}}-{{u}^{2}}=2as$

Therefore, $-{{35}^{2}}=2a\left( 200 \right)$

$a=-3.0625m/{{s}^{2}}$

From first equation of motion, time (t) taken by the car to stop can be obtained as:

$v=u+at$

$0=35-\left( 3.065 \right)t$

$t=11.44s$

Hence, the retardation of the car (assumed uniform) is $3.0625m/{{s}^{2}}$, and it takes $11.44s$ for the car to stop.

12. 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

Ans: The above statement is true. When an object is thrown vertically up in the air, its speed becomes zero at maximum height. It has acceleration equal to the acceleration due to gravity (g) Which acts in the downward direction at that point.

b) With zero speed may have non-zero velocity,

Ans: The above statement is false as speed is the magnitude of velocity. If speed is zero, the magnitude of velocity along with the velocity is zero.

c) With constant speed must have zero acceleration,

Ans: The above statement is true. If a car is moving on a straight highway with constant speed, it will have a constant velocity. Acceleration is defined as the rate of change of velocity. Hence, the acceleration of the car is also zero.

d) With a positive value of acceleration must be speeding up.

Ans: The above statement is false. If acceleration is positive and velocity is negative at the instant time is taken as origin. Thus, for all the time before velocity becomes zero, there is a slowing down of the particle. This case occurs when a particle is projected upwards. This statement will be true when both velocity and acceleration are positive, at that instant time taken as origin. This case happens when a particle is moving with positive acceleration or falling vertically downwards from a height.

13. Suggest a suitable physical situation for each of the following graphs (figure):

(Image Will Be Updated Soon)

Ans: Consider the figure given in the question:

a) From the x-t graph given it is clear that initially, a body was at rest. Further, its velocity increases with time and attains an instantaneous constant value. The velocity then reduces to zero with an increase in time. Further, its velocity increases with time in the opposite direction and acquires a constant value. A similar physical situation arises when a football (initially kept at rest) is kicked and gets rebound from a rigid wall so that its speed gets reduced. Then, it passes from the player who has kicked it and ultimately stops after some time.

b) From the given v-t graph it is clear that the sign of velocity changes and its magnitude decreases with the passage of time. This type of situation arises when a ball is dropped on the hard floor from a height. It strikes the floor with some velocity and upon rebound, its velocity decreases by a factor. This continues till the velocity of the ball eventually becomes zero.

c) From the given a-t graph it is clear that initially, the body is moving with a certain uniform velocity. Its acceleration increases for a short interval of time, which again drops to zero. This shows that the body again starts moving with the same constant velocity. This type of physical situation arises when a hammer moving with a uniform velocity strikes a nail.

14. 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 in the figure. Choose the correct entries in the brackets below;

(Image Will Be Updated Soon)

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

Ans: From the graph, it is clear that $OP<OQ$. Therefore, A lives closer to the school than B.

b) (A/B) starts from the school earlier than (B/A)

Ans: From the graph it is clear that for  $ x\text{ }=\text{ }0,\text{ }t\text{ }=\text{ }0 $  for A and t has some finite value for B. Hence, A starts from the school earlier than B.

c) (A/B) walks faster than (B/A)

Ans: As the velocity is equal to the slope of the x-t graph, in the case of uniform motion and slope of the x-t graph for B is greater than that for A. Thus, B walks faster than A.

d) A and B reach home at the (same/different) time

Ans: From the graph, it is clear that both A and B reach their respective homes at the same time.

e) (A/B) overtakes (B/A) on the road (once/twice).

Ans: As B moves later than A and his/her speed is greater than that of A. From the graph, it is clear that B overtakes A only once on the road.

4 Marks Questions

1. On a two-lane road, car A is travelling with a speed of  $ 36\text{ }km/hr $. Two cars B and C approach car A in opposite directions with a speed of  $ 54\text{ }km/hr $  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?

Ans: In the above question it is given that:

Velocity of car A is  $ 36\text{ }km/hr=10m/s $ . $ Velocity\text{ }of\text{ }car\text{ }B\text{ }=\text{ }Velocity\text{ }of\text{ }car\text{ }C=54km/hr=15m/s $ 

Relative velocity of car A with respect to car B,

= 15 – 10 = 5 m/s

Relative velocity of car A with respect to car C,

$ =\text{ }15\text{ }+\text{ }10\text{ }=\text{ }25\text{ }m/s $ 

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

Hence,  $ s\text{ }=\text{ }1\text{ }km\text{ }=\text{ }1000\text{ }m $ .

Time taken (t) by car C to cover  $ 1000\text{ }m $  (i.e., to overtake A)  $ =\text{ }1000\text{ }/\text{ }25\text{ }=\text{ }40\text{ }s $ .

Thus, to avoid an accident, car B must cover the same distance in a maximum of  $ 40\text{ }s $ .

Using second equation of motion, minimum acceleration (a) produced by car B will be:

$s=ut+\left( 1/2 \right)a{{t}^{2}}$

$1400=\left( 15\times 40 \right)+\left( \dfrac{1}{2}\times a\times {{\left( 40 \right)}^{2}} \right)$

$\Rightarrow a=1m/{{s}^{2}}$

Hence, the minimum acceleration of car B required to avoid an accident is $1m/{{s}^{2}}$.

2. Explain clearly, with examples, the distinction between:

a) The 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;

Ans: The shortest distance (which is a straight line) between the initial and final positions of the particle gives the magnitude of displacement over an interval of time. The total path length of a particle is the actual path length covered by the particle in a given interval of time. For 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. Then, the magnitude of displacement of the particle is AC.

(Image Will Be Updated Soon)

Whereas, total path length  $ =\text{ }AB\text{ }+\text{ }BC $ 

We know that the magnitude of displacement can never be greater than the total path length. But, in some cases, both quantities are equal to each other.

b) The 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).

Ans: We know that: 

$ Magnitude\text{ }of\text{ }average\text{ }velocity\text{ }=\text{ }Magnitude\text{ }of\text{ }displacement\text{ }/\text{ }Time\text{ }interval $ 

Hence, for the given particle,

Average velocity $=AC/t$

$ Average\text{ }speed\text{ }=\text{ }Total\text{ }path\text{ }length\text{ }/\text{ }Time\text{ }interval $ 

 $ =\left( AB+BC \right)/t $ 

Since,  $ AB+BC>AC $, the average speed is greater than the magnitude of average velocity. The two quantities will be equal if the particle continues to move along a straight line.

3. 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 Chapter14). Give the signs of position, velocity, and acceleration variables of the particle at $t=0.3s,1.2s,-1.2s$.

(Image Will Be Updated Soon)

Ans: The required signs are:

Negative, Negative, Positive

Positive, Positive, Negative

Negative, Positive, Positive

When a particle executes simple harmonic motion (SHM), acceleration (a) is given by the relation: $a=-{{\omega }^{2}}x$

Where,

$\omega $ is the angular frequency ... 

i) $ ~t\text{ }=\text{ }0.3\text{ }s $ 

For this time interval, x is negative. Hence, the slope of the x-t plot will be negative. Thus, both position and velocity are negative. However, using equation (i) the acceleration of the particle will be positive.

ii) $ ~t\text{ }=1.2s $ 

For this time interval, x is positive. Hence, the slope of the x-t plot will be positive. Thus, both position and velocity are positive. However, using equation (i)the acceleration of the particle comes to be negative.

iii) $ \text{t }=-1.2s $ 

For this time interval, x is negative. Hence, the slope of the x-t plot will be negative. Thus, both x and t are negative, the velocity comes to be positive. From equation (i), it can be interpreted that the acceleration of the particle will be positive.

4. On a long horizontally moving belt (figure), a child runs to and fro with a speed  $ 9\text{ }km/hr $  (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\text{ }km\text{ }{{h}^{1}} $.

(Image Will Be Updated Soon)

For an observer on a stationary platform outside, what is the

a) Speed of the child running in the direction of motion of the belt?

Ans: As the boy is running in the same direction of the motion of the belt, his speed (as observed by the stationary observer) will be:

$ {{v}_{bB\text{ }}}=\text{ }{{v}_{b}}\text{ }+\text{ }{{v}_{B}}\text{ }=\text{ }9\text{ }+\text{ }4\text{ }=\text{ }13\text{ }km/h $.

b) Speed of the child running opposite to the direction of motion of the belt?

Ans: As 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:

$ {{v}_{bB\text{ }}}=\text{ }{{v}_{b}}\text{ }+\text{ }\left( -{{v}_{B}}\text{ } \right)=\text{ }9-4\text{ }=\text{ 5}km/h $

c) Time taken by the child in (a) and (b)? Which of the answers alter if motion is viewed by one of the parents?

Ans: We know that: 

$ Distance\text{ }between\text{ }the\text{ }child’s\text{ }parents\text{ }=\text{ }50\text{ }m $ 

Since, 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\text{ }km/h\text{ }=\text{ }2.5\text{ }m/s $ . Thus, the time taken by the child to move towards one of his parents is  $ 50/2.5\text{ }=\text{ }20s $ . If the motion is viewed by any one of the parents, answers 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\text{ }km/h $. Therefore, it can be interpreted that the time taken by the child to reach any one of his parents remains unaltered.

5. Two towns A and B are connected by regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of  $ 20\text{ }km/hr $  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?

Ans: In the above question it is given that:

Consider V to be the speed of the bus running between towns A and B.

Speed of the cyclist is  $ v=20\text{ }km/hr $ .

The relative speed of the bus moving in the direction of the cyclist will be $V-v=\left( V-20 \right)m/s$.

The bus went past the cyclist every  $ 18 $  min i.e.,  $ 18\text{ }/\text{ }60\text{ }h $  (when he moves in the direction of the bus).

Hence, distance covered by the bus  $ =\text{ }\left( V\text{ }-\text{ }20 \right)\text{ }\times \text{ }18\text{ }/\text{ }60\text{ }km $ ...... (i)

As one bus leaves after every T minutes, the distance travelled by the bus will be  $ =V\text{ }\times \text{ }T\text{ }/\text{ }60 $ ...... (ii)

As equations (i) and (ii) are equal.

$ \left( V\text{ }-\text{ }20 \right)\text{ }\times \text{ }18\text{ }/\text{ }60\text{ }=\text{ }VT\text{ }/\text{ }60 $ ...... (iii)

Relative speed of the bus moving in the opposite direction of the cyclist will be  $ \left( V\text{ }+\text{ }20 \right)\text{ }km/h $ .

Thus, time taken by the bus to go past the cyclist  $ =\text{ }6\text{ }min\text{ }=\text{ }6\text{ }/\text{ }60\text{ }hr $ 

$ \Rightarrow \left( V\text{ }+\text{ }20 \right)\times 6\text{ }/\text{ }60\text{ }=\text{ }VT\text{ }/\text{ }60 $ ...... (iv)

From (iii) and (iv), we get

$ \left( V\text{ }+\text{ }20 \right)\text{ }\times \text{ }6\text{ }/\text{ }60\text{ }=\text{ }\left( V\text{ }-\text{ }20 \right)\text{ }\times \text{ }18\text{ }/\text{ }60 $ 

$ V\text{ }+\text{ }20\text{ }=\text{ }3V\text{ }\text{ }60 $ 

$ 2V\text{ }=\text{ }80 $ 

$ V\text{ }=\text{ }40\text{ }km/h $ , which is the required speed.

Substituting the value of V in equation (iv), 

$ \left( 40\text{ }+\text{ }20 \right)\text{ }\times \text{ }6\text{ }/\text{ }60\text{ }=\text{ }40T\text{ }/\text{ }60 $ 

$ T\text{ }=\text{ }360\text{ }/\text{ }40\text{ }=\text{ }9\text{ }min $ , which is the required time period.

6. 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?

Ans: From the graph given in the question,

Average acceleration is greatest in interval  $ 2 $ .

Average speed is greatest in intervals of $3$.

v is positive for intervals  $ 1,\text{ }2 $ , and $3$.

a is positive for intervals $1$ & $3$ and negative in interval $2$

$ a\text{ }=\text{ }0 $  at A, B, C, D

Acceleration is calculated as the slope of the speed-time graph. In the given case, it is given by the slope of the speed-time graph within the given interval of time.

As the slope of the given speed-time graph is maximum in interval  $ 2 $, average acceleration will be the greatest in this interval.

From the time-axis, the height of the curve gives the average speed of the particle. It is clear that the height is the greatest in interval 3. Thus, the average speed of the particle is the greatest in the interval $3$.

For interval $1$:

The slope of the speed-time graph is positive. Hence, acceleration is positive. Similarly, the speed of the particle is positive in this interval.

In interval $2$:

As the slope of the speed-time graph is negative, acceleration is negative in this interval. However, speed is positive because it is a scalar quantity.

In interval $3$:

As the slope of the speed-time graph is zero, acceleration is zero in this interval.

However, here the particle acquires some uniform speed. It is positive in this interval.

Points A, B, C, and D are all parallel to the time axis. Thus, the slope is zero at these points.

Therefore, at points A, B, C, and D, the acceleration of the particle is zero.

5 Marks Questions

1. Velocity time graph of a moving particle is shown. Find the displacement (1) 0 – 4 s (2) 0 – 8 (3) 0 12 s from the graph. Also, write the differences between distance and displacement.

(Image Will Be Updated Soon)

Ans: Considering the graph given in the question,

(1) Displacement for the interval  $ \left( 0\text{ }\text{ }4 \right)\text{ }s $ :

$ {{S}_{1}}\text{ }=\text{ }area\text{ }of\text{ }OABD $ 

$ {{S}_{1}}\text{ }=\text{ }15\times 4\text{ }=\text{ }60\text{ }m $ 

(2) Displacement for the interval  $ \left( 0\text{ }\text{ }8 \right)s $ :

$ {{S}_{2}}\text{ }=\text{ }{{S}_{1}}\text{ }+\text{ }area\text{ }\left( CDEF \right) $

$ {{S}_{2}}\text{ }=\text{ }60\text{ }+\text{ }\left( -5 \right)\text{ }4\text{ }=\text{ }60\text{ }-20\text{ }=\text{ }40m $

(3) Displacement for the interval   $ \left( 0\text{ }\text{ }12 \right)s $ :

$ {{S}_{3}}\text{ }=\text{ }{{S}_{1\text{ }}}+\text{ }area\text{ }\left( CDEF \right)\text{ }+\text{ }area\text{ }\left( FGHI \right) $ 

$ {{S}_{3\text{ }}}=\text{ }60\text{ }\text{ }20\text{ }+\text{ }40\text{ }=\text{ }80m $ 

Differences between distance and displacement are:

2. A player throws a ball upwards with an initial speed of  $ 29.4\text{ }m/s $.

a) What is the direction of acceleration during the upward motion of the ball?

Ans: Acceleration of the ball (which is actually acceleration due to gravity) always acts in the downward direction towards the center of the Earth, irrespective of the direction of the motion of the ball.

b) What are the velocity and acceleration of the ball at the highest point of its motion?

Ans: 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., $ g\text{ }=\text{ }9.8\text{ }m/{{s}^{2}} $ . At maximum height, velocity of the ball becomes zero.

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 the x-axis, and give the signs of position, velocity, and acceleration of the ball during its upward, and downward motion.

Ans: The sign of position is positive, a sign of velocity is negative, and a sign of acceleration is positive during upward motion. During downward motion, the signs of position, velocity, and acceleration are all positive.

d) To what height does the ball rise and after how long does the ball return to the player’s hands? (Take  $ g\text{ }=\text{ }9.8\text{ }m/{{s}^{2}} $  and neglect air resistance).

Ans: Initial velocity of the ball,  $ u\text{ }=\text{ }29.4\text{ }m/s $ .

Final velocity of the ball,  $ v\text{ }=\text{ }0 $  (At maximum height, the velocity of the ball becomes zero)

Acceleration,  $ a=g\text{ }=9.8\text{ }m/{{s}^{2}} $ 

From first equation of motion, 

$v=u+at$

$0=-29.4+\left( 9.8 \right)t$

$t=\dfrac{29.4}{9.8}=3s$

$ Time\text{ }of\text{ }ascent\text{ }=\text{ }Time\text{ }of\text{ }descent $ 

Hence, the total time taken by the ball to return to the player’s hands $ =\text{ }3\text{ }+\text{ }3\text{ }=\text{ }6\text{ }s $.

3. 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  $ \mathbf{t}\text{ }=\text{ }\mathbf{0}\text{ }\mathbf{to}\text{ }\mathbf{12}\text{ }\mathbf{s} $ .

Ans: In the above question it is given that:

Ball is dropped from a height of  $ s\text{ }=\text{ }90\text{ }m $ .

Initial velocity of the ball is  $ u\text{ }=\text{ }0 $ .

Acceleration is $a=g=9.8m/{{s}^{2}}$.

Consider,

Final velocity of the ball to be v.

Using second equation of motion, time (t) taken by the ball to hit the ground can be obtained

as:

$s=ut+\left( 1/2 \right)a{{t}^{2}}$

$90=0+\left( 1/2 \right)9.8{{t}^{2}}$

$t=\sqrt{18.38}=4.29s$

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

$v=u+at$

$v=0+9.8\left( 4.29 \right)=42.04m/s$

Rebound velocity of the ball is calculated as:

${{u}_{r}}=9v/10=9\left( \dfrac{42.04}{10} \right)=37.84m/s$

Time (t) taken by the ball to reach maximum height is obtained with the help of first equation of motion as:

$v={{u}_{r}}+at'$

$0=37.84+\left( -9.8 \right)t'$

$t'=37.84/9.8=3.86s$

Total time taken by the ball will be $t+t'=4.29+3.86=8.15s$.

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

The velocity with which the ball rebounds from the floor will be ${{u}_{r}}=9v/10=9\left( \dfrac{37.84}{10} \right)=34.05m/s$

The total time taken by the ball for the second rebound will be $t+t'=8.15+3.86=12.01s$

The speed-time graph of the ball is represented in the given figure as:

(Image Will Be Updated Soon)

4. A man walks on a straight road from his home to a market 2.5 km away with a speed of  $ 5\text{ }km/hr $. Finding the market closed, he instantly turns and walks back home with a speed of  $ 7.5\text{ }km/hr $. What is the

(a) the magnitude of average velocity, and

(b) the 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 the magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero)

Ans: In the above question it is given that:

Time taken by the man to reach the market from home is ${{t}_{1}}=2.5/5=1/2hr=30\min $.

Time taken by the man to reach home from the market is ${{t}_{2}}=2.5/7.5=1/3hr=20\min $. 

Total time taken in the whole journey  $ =\text{ }30\text{ }+\text{ }20\text{ }=\text{ }50\text{ }min $ .

i) 0 to 30 min

$ Average\text{ }velocity\text{ }=\text{ }Displacement/Time $ 

$ Average\text{ }speed\text{ }=\text{ }Distance/Time $ 

ii) 0 to 50 min

$ Time\text{ }=\text{ }50\text{ }min\text{ }=\text{ }50/60\text{ }=\text{ }5/6\text{ }h $ 

$ Net\text{ }displacement\text{ }=\text{ }0 $ 

$ Total\text{ }distance\text{ }=\text{ }2.5\text{ }+\text{ }2.5\text{ }=\text{ }5\text{ }km $ 

$ Average\text{ }velocity\text{ }=\text{ }Displacement\text{ }/\text{ }Time\text{ }=\text{ }0 $ 

$ Average\text{ }speed\text{ }=\text{ }Distance\text{ }/\text{ }Time\text{ }=\text{ }5/\left( 5/6 \right)\text{ }=\text{ }6\text{ }km/h $ 

iii) 0 to 40 min

$ Speed\text{ }of\text{ }the\text{ }man\text{ }=\text{ }7.5\text{ }km/h $ 

$ Distance\text{ }travelled\text{ }in\text{ }first\text{ }30\text{ }min\text{ }=\text{ }2.5\text{ }km $ 

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

$ =\text{ }7.5\text{ }\times \text{ }10/60\text{ }=\text{ }1.25\text{ }km $ 

$ Net\text{ }displacement\text{ }=\text{ }2.5\text{ }\text{ }1.25\text{ }=\text{ }1.25\text{ }km $ 

$ Total\text{ }distance\text{ }travelled\text{ }=\text{ }2.5\text{ }+\text{ }1.25\text{ }=\text{ }3.75\text{ }km $ 

$ Average\text{ }velocity\text{ }=\text{ }Displacement\text{ }/\text{ }Time\text{ }=\text{ }1.25\text{ }/\text{ }\left( 40/60 \right)\text{ }=\text{ }1.875\text{ }km/h $ 

$ Average\text{ }speed\text{ }=\text{ }Distance\text{ }/\text{ }Time\text{ }=\text{ }3.75\text{ }/\text{ }\left( 40/60 \right)\text{ }=\text{ }5.625\text{ }km/h $.

5. A three-wheeler starts from rest, accelerates uniformly with  $ 1\text{ }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 nth second (n = 1,2,3....) versus n. What do you expect this plot to be during accelerated motion: a straight line or a parabola?

Ans: In the above question it is given that:

$ Initial\text{ }velocity=u=0m/s $ 

$ Acceleration=a=1m/{{s}^{2}} $ 

Distance covered by a body in nth second is given by the relation

${{s}_{n}}=u+\left( 2n-1 \right)a/2$

Hence,

${{s}_{n}}=\left( 2n-1 \right)/2=n-1/2$ …… (1)

In this case, this relation shows that: ${{s}_{n}}$ linearly varies with n.

Now, substituting different values of n in equation (1), we get the following table:

The plot between n and will be a straight line shown in below figure:

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6. Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of $15m/s$ and $30m/s$. Verify that the graph shown in the 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\text{ }=\text{ }10m/{{s}^{2}} $ . Give the equations for the linear and curved parts of the plot.

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Ans: In the above question it is given that:

Initial velocity ${{u}_{1}}=15m/s$.

Acceleration, $a=-g=-10m/{{s}^{2}}$.

Using the relation,

${{x}_{1}}={{x}_{0}}+{{u}_{1}}t+\left( 1/2 \right)a{{t}^{2}}$

Where, height of the cliff, ${{x}_{0}}=200m$.

${{x}_{1}}=200+15t-5{{t}^{2}}$ ….. (1)

When this stone hits the ground, ${{x}_{1}}=0m$.

$\therefore -5{{t}^{2}}+15t+200=0$

${{t}^{2}}-3t-40=0$

${{t}^{2}}-8t+5t-40=0$

$\left( t-8 \right)\left( t+5 \right)=0$

$\therefore t=8,-5$

Hence, $t=8s$.

For second stone:

Initial velocity, ${{u}_{2}}=30m/s$.

Acceleration, $a=-g=-10m/{{s}^{2}}$.

Using the relation,

${{x}_{2}}={{x}_{0}}+{{u}_{2}}t+\left( 1/2 \right)a{{t}^{2}}$

Where, height of the cliff, ${{x}_{0}}=200m$.

${{x}_{2}}=200+30t-5{{t}^{2}}$ ….. (2)

When this stone hits the ground, ${{x}_{2}}=0m$.

$\therefore -5{{t}^{2}}+30t+200=0$

${{t}^{2}}-6t-40=0$

${{t}^{2}}-10t+4t-40=0$

$\left( t-10 \right)\left( t+4 \right)=0$

$\therefore t=10,-4$

Hence, $t=10s$.

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

${{x}_{2}}-{{x}_{1}}=\left( 200+30t-5{{t}^{2}} \right)-\left( 200+15t-5{{t}^{2}} \right)$

${{x}_{2}}-{{x}_{1}}=15t$ …… (3)

Equation (3) represents the linear path of both stones. Due to this linear relation between ${{x}_{2}}-{{x}_{1}}$ and t, the path remains a straight line till  $ 8\text{ }s $ .

${{\left( {{x}_{2}}-{{x}_{1}} \right)}_{\max }}=15\times 8=120m$

This is in accordance with the given graph.

After 8 s, only a second stone is in motion whose variation with time is given by the quadratic equation: ${{x}_{2}}-{{x}_{1}}=200+30t-5{{t}^{2}}$.

Hence, the equation of linear path is ${{x}_{2}}-{{x}_{1}}=15t$ and curved path is ${{x}_{2}}-{{x}_{1}}=200+30t-5{{t}^{2}}$.

7. The speed-time graph of a particle moving along a fixed direction is shown in the figure. Obtain the distance traversed by the particle between $\left( a \right)\text{0s to 10s}$, (b) 2s to 6s. What is the average speed of the particle over the intervals in (a) and (b)?

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Ans: From the graph given in the question:

(a)  $ Distance\text{ }travelled\text{ }by\text{ }the\text{ }particle\text{ }=\text{ }Area\text{ }under\text{ }the\text{ }given\text{ }graph $ 

$=\left( 1/2 \right)\times \left( 10-0 \right)\times \left( 12-0 \right)=60m$

$ Average\text{ }speed=Distance/Time=60/10=6m/s $ 

(b) Let  $ {{s}_{1}} $  and  $ {{s}_{2}} $  be the distances covered by the particle between time  $ t\text{ }=\text{ }2\text{ }s $  to  $ 5\text{ }s $  and  $ t\text{ }=\text{ }5\text{ }s $  to  $ 6\text{ }s $  respectively.

Total distance (s) covered by the particle in time  $ t\text{ }=\text{ }2\text{ }s $  to  $ 6\text{ }s $  will be:

$s={{s}_{1}}+{{s}_{2}}$ …… (i)

For distance  $ {{s}_{1}} $ :

Let u′ be the velocity of the particle after  $ 2\text{ }s $  and a’ be the acceleration of the particle for $t=0\text{s to 5s}$.

Since the particle undergoes uniform acceleration in the interval $t=0\text{s to 5s}$, from first equation of motion, acceleration can be obtained as:

$v=u+at$

Where,

$ v\text{ }=\text{ }Final\text{ }velocity\text{ }of\text{ }the\text{ }particle $ 

$a'=12/5=2.4m/{{s}^{2}}$

Again, from first equation of motion, we have

$v=u+at$

$=0+2.4\times 2=4.8m/s$

Distance travelled by the particle between time  $ 2\text{ }s $  and  $ 5\text{ }s $  i.e., in  $ 3\text{ }s $  will be:

${{s}_{1}}=u't+\left( 1/2 \right)a'{{t}^{2}}$

$=4.8\left( 3 \right)+\left( 1/2 \right)\times 2.4\times {{3}^{2}}=25.2m$ …… (ii)

For distance ${{s}_{2}}$,

Let a’’ be the acceleration of the particle between time $\text{t=5s and t=10s}$.

From first equation of motion,

$v=u+at$

$0=12+a''\times 5$

$a''=-2.4m/{{s}^{2}}$

Distance travelled by the particle in 1s (i.e.,  $ t\text{ }=\text{ 5}s $  to  $ 6s $ )

${{s}_{2}}=u''t+\left( 1/2 \right)a''{{t}^{2}}$

$=12\times 1+\left( 1/2 \right)\left( -2.4 \right)\times {{\left( 1 \right)}^{2}}=10.8m$ …… (iii)

From equations (i), (ii), and (iii), we get:

$ \text{s=25}\text{.2+10}\text{.8=36m} $ 

Hence,  $ \text{Average speed = 36 / 4 = 9 m/s} $.

Important Questions of Chapter 13 Class 11 Physics - Free Pdf Download

Students who are weak in physics and do not have a strong core knowledge might find the chapter motion in a straight line difficult. The chapter is full of theories and concepts which can be difficult to grasp and it will be a hindrance for them to achieve good marks. Students need to try and avoid this problem and score good marks in the subject. One of the best ways to overcome such a problem is to study and practice. Students should take time and solve the class 11 physics chapter 3 important questions, an exercise which will help students get the marks they deserve. Students should study theories as well as practice the practical aspects of the chapter. Physics Class 11 chapter 3 important questions is a section which will help students test their knowledge of the chapter. This will make the students regular and efficient to write the exam. 

Students should take the help of physics class 11 chapter 3 important questions after they finish studying the chapter. It's going to be helpful and useful for them. The important questions for class 11 physics chapter 3 are available on the Vedantu site for free. The downloaded CBSE class 11 chapter 3 important questions are an important guide for all the students of class 11.

Class 11 Physics Chapter 3 Important Questions

Students will learn about the concepts and theories of motion in a straight line when they study chapter 3 of class 11 in physics. It's an activity which is going to help students understand and gain knowledge about the important questions of chapter 3 physics class 11. Some of the knowledge that the students will learn is as follows:

Motion: One of the most important topics in the chapter of motion in a straight line is motion. Everything in the solar system moves. It is said that an object is in motion if it changes its position from place to another with time. Motion is a relative concept and a body is always in relative motion if it is changing positions. There are two main branches in physics which teach the concept of motion. The two concepts are as follows: 

• Kinematics - Kinematics is the concept which describes the motion of objects. It doesn't look at the cause of the motion.

• Dynamics - Dynamics is the concept which describes the motion of the object to the force which is causing the motion.

Point Object -  Point objects refer to the length covered by objects. When the length covered by an object is more or larger in comparison to the size of the object, the object is said to be a point object. An example of a point object is when the railway carriage goes from point a to b without any jerk. 

Total Path Length - Total Path Length refers to the distance covered by an object. It refers to the length of actual path traversed between a particular object in motion. The measurement of the distance is from the initial point to the final position of the object. 

Types of Motion - An important topic which is the part of the motion in a straight line is the type of motion. It is important to know the specific position of an object. We need to know the coordinate positions of the object. The number of coordinate positions may vary between different kinds of objects. Sometimes there are two coordinates or three coordinates. Here is the classification of motion based on different coordinates: 

• One-Dimensional Motion: When an object moves in a straight line or along a path is said to be in one-dimensional motion. An example of one-dimensional motion is a freely falling body under gravity. 

• Two-Dimensional Motion: An object moves in a plane undergoes two-dimensional motion. An example of two-dimensional motion is carrom board coins. 

• Three-Dimensional Motion: An object moving in space undergoes three-dimensional motion. An example of three-dimensional motion is the motion of an aeroplane. 

Displacement: It is an important topic under the topic of motion in a straight line. Displacement is an object at a given time or the change in the position of the direction of the object during that time. It is a vector drawn from its starting position to the last position. 

Uniform Speed and Uniform Velocity: An object is said to be moving at an uniform speed if it covers equal distances in equal intervals of time. The intervals of time can be small or big but must be equal. 

An object is said to be moving in uniform velocity if it covers an equal amount of displacement in equal intervals of time. The intervals of time can be small or big but must be equal. 

Variable Speed and Variable Velocity: An object is said to move with variable speed if it covers unequal distances in equal intervals of time. The time interval can be small or large. 

An object is said to be moving with variable velocity if it covers an unequal amount of displacement in equal intervals of time. The intervals of time can be small or big.

Important Questions for Motion in a Straight Line Class 11

Students can refer to the chapter 3 physics class 11 important questions to learn the concepts of motion in a straight line. We have enlisted important and potential questions that are most likely to appear in the examination: 

• When is the condition of the relationship s= ut correct?

• If two balls of different masses are thrown vertically upwards with alike initial speed. Which one will rise to a higher height.  

• Under what circumstances is the relative velocity of two bodies having equal velocity?

• A railway train 450m is going from Chennai railway station to Pondicherry. Can we consider the railway train as a point object?

• Lepakshi went from her house to college 3.5kms away. On finding her college closed, she returned home immediately. What is her net displacement and the total distance covered by her? 

• Justify the sentence: speed of an object be negative. 

• Explain the following examples of motion. Justify if the body can be considered as a point object. 

• A railway carriage moving from one station to another without any jerks to stop the train. 

• A monkey sitting on a top of a woman cycling smoothly on a circular track. 

• Under what conditions are the shape of the displacement time graph for uniform linear motion?

• Draw a displacement time graph with three points showing the different signs of velocities on the graph.
  
  

Benefits of Important Questions For Class 10 Science Chapter 3

Students who will give their class 11 examinations get a lot of advantage from class 11 Physics chapter 3 important questions. These important questions are going to help students score good marks and excel in their board exams. Some of the benefits of the class 11 science ch 3 important questions are as follows:

• The Vedantu team does proper research before making the list of class 11 physics motion in a straight line important questions so that the questions included in the list have the highest probability of coming in the exams.

• There are experts at Vedantu who review these questions, who are professionals in the field and their experience and expertise ensure that no mistake is committed. 

• The class 11 physics motion in a straight line important questions are listed keeping in mind the format issued by the CBSE board; thus, it prepares students for the questions that they can expect in their physics exam. 

• The solutions to these important questions are given in a way that is detailed and explanative for all students to understand and learn.
  
  

Conclusion

Students who find physics difficult and are weak in the subject and are unable to understand chapter 3 must consider using class 11 physics motion in a straight line important questions as a reference. It is going to be helpful for the students and they will be able to score the highest marks in the board exams. Referring to the important questions for class 11 physics chapter 3 is going to improve the students’ confidence in excelling the exam.

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