# Free NCERT Solutions For Class 11 Physics Chapter 3 : Motion in a Straight Line Free PDF Download

**Question 1:In which of the following examples of motion, can the body be considered approximately apoint 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.**

Solution 1:

(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.

**Question 2: 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 figure. Choose the correct entries in the bracketsbelow;(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).**

Solution 2:

(a) As OP OQ , A lives closer to the school than B.

(b) For x = 0, t = 0 for A; while t has some finite value for B. Therefore, A starts from the school earlier than B.

(c) Since the velocity is equal to slope of x-t graph in case of uniform motion and slope of x-t graph for B is greater that that for A =, hence B walks faster than A.

(d) It is clear from the given graph that both A and B reach their respective homes at the same time.

(e) 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.

**Question 3: 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/h. Choose suitable scales and plot the x-t graph of her motion**

**Ans:**

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

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

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

**Question 4: A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followedagain 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:**

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

Net distance covered = 5 – 3 = 2 m

Net time taken to cover 2 m = 8 s

Drunkard covers 2 m in 8 s.

Drunkard covered 4 m in 16 s.

Drunkard covered 6 m in 24 s.

Drunkard covered 8 m in 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

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

**Question 5: 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 ?**

Solution 5:

Speed of the jet airplane, = 500 km/h

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

Speed of its products of combustion with respect to the ground

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

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

**Question 6: 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?**

Solution 6:

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, a can be calculated as:

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

v=u+at

t=(v-u)/a=(-35)/(-3.06)=11.44s

**Question 7: 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 ^{-2}. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them ?**

**Ans:**

For train A:

Initial velocity, u = 72 km/h = 20 m/s

Time, t= 50s

Acceleration, a_{I}= 0 (Since it is moving with a uniform velocity)

From second equation of motion, distance(s_{1}) covered by train A can be obtained as:

**Question 11: Read each statement below carefully and state with reasons and examples, if it is true orfalse; 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.**

**Ans:**

(a) True, when an object is thrown vertically up in the air, its speed becomes zero at maximum height. However, it has acceleration equal to the acceleration due to gravity (g) that acts in the downward direction at that point.

(b) False, speed is the magnitude of velocity. When speed is zero, the magnitude of velocity along with the velocity is zero.

(c) True, a car moving on a straight highway with constant speed will have constant velocity.

Since acceleration is defined as the rate of change of velocity, acceleration of the car is also

zero.

(d) False, this statement is false in the situation when acceleration is positive and velocity is negative at the instant time taken as origin. 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.

This statement is true when both velocity and acceleration are positive, at the instant time taken as origin. Such a case happens when a particle is moving with positive acceleration or falling vertically downwards from a height.

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

**Ans:**Ball is dropped from a height, s = 90 m

Initial velocity of the ball, u = 0

Acceleration, a=g=9.8ms

^{-2}

Final velocity of the ball = v

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

as:

**Question 13: Explain clearly, with examples, the distinction between:(a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;(b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first.When is the equality sign true? [For simplicity, consider one-dimensional motion only].**

Solution 13:

(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 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 = AC.

Whereas, total path length = AB + BC

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 to each other.

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

For the given particle,

Average velocity =

AC t /

Average speed = Total path length / Time interval

/ AB BC t

Since

AB BC AC ,

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.

**Question 15: In Exercises 13 and 14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speedand magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?**

**Ans:**

Instantaneous velocity is given by the first derivative of distance with respect to time i.e. , Here, the time interval is so small that it is assumed that the particle does not change its direction of motion. As a result, both the total path length and magnitude of displacement become equal is this interval of time.

Therefore, instantaneous speed is always equal to instantaneous velocity.

**Question 20: Figure gives the x-t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter14). Give the signs of position,velocity and acceleration variables of the particle at t=0.3s,1.2s,-1.2s**

**Ans:**

Negative, Negative, Positive

Positive, Positive, Negative

Negative, Positive, Positive

For simple harmonic motion (SHM) of a particle, acceleration (a) is given by the relation:

angular frequency … (i) t = 0.3 s

In this time interval, x is negative. Thus, the slope of the x-t plot will also be negative.

Therefore, both position and velocity are negative. However, using equation (i), acceleration of the particle will be positive.

In this time interval, x is positive. Thus, the slope of the x-t plot will also be positive.

Therefore, both position and velocity are positive. However, using equation (i), acceleration of

the particle comes to be negative.

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

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

**Ans:**

Interval 3 (Greatest), Interval 2 (Least)

Positive (Intervals 1 & 2), Negative (Interval 3)

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

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

**Question 28: 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 t1 to t2 :**

**Solution 28:**

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.

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