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# Class 11 Physics NCERT Solutions of chapter 3-Motion in a Straight Line

Unlock the pathway to success in Class 11 Physics with comprehensive NCERT Solutions tailored for Chapter 3: Motion in a Straight Line. Crafted by subject matter experts, these solutions serve as invaluable aids for CBSE Board exam and class test preparations.

Class 11 Physics NCERT Solutions for Chapter 3 delve deep into the intricacies of Motion in a Straight Line, elucidating key concepts with clarity and precision. Whether you’re a diligent student aiming for academic excellence or gearing up for competitive exams like NEET, engineering entrance exams, NDA, or CDA, these solutions are your ultimate companion.

Designed in accordance with CBSE guidelines, these solutions provide a solid foundation for understanding the nuances of Chapter 3, enabling you to grasp complex concepts effortlessly. Elevate your learning experience and conquer the challenges of Motion in a Straight Line with Class 11 Physics NCERT Solutions.

Q1 In which of the following examples of motion, can the body be considered
approximately a point object:
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table.

Ans.(a) A railway carriage moving without jerks between two stations is considered as a point object as the distance traveled by train between two stations is very large as compared with the size of the train.
(b) A monkey sitting on top of a man cycling smoothly on a circular track is considered as a point object if the radius of the circular track through which the cyclist has traveled is very large as compared to the monkey sitting on top of the cyclist.
(c) A spinning cricket ball that turns sharply on hitting the ground can not be considered a point object because the distance which causes the ball to spin on hitting the ground is not much as compared to the size of the ball.
(d) A tumbling beaker that has slipped off the edge of a table can not be considered a point object since the distance traveled by the beaker from the table to the ground is not very large as compared to the size of the beaker.

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Q 3. 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 Fig. Choose the correct entries in the brackets below:
(a) (A/B) lives closer to the school than (B/A).
(b) (A/B) starts from 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).

Ans.

(a) A lives closer to the school than B because OP<OQ where OP and OQ are the distances of the homes of A and B from the school(O) respectively.
(b) A starts from school earlier than B because according to the graph A starts at t=0 while B starts after time t.
(c) B walks faster than A because the slope of graph B is more than the slope of graph A, as we know the speed is equal to the slope in the x-t graph.
(d) Let P reaches to his home after tp time and Q reaches after tq time

Time is taken by A to reach home = tp and time taken by B to reach home is tq, therefore A and B reach home at different time.

(e) B overtakes A on the road once because in the given x-t graph the graph of A and B intersect at one point and the speed of B is more than that of A.

Question 3. 3. A woman starts from her home at 9.00 am, walks with a speed of 5 km h-1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h-1. Choose suitable scales and plot the x-t graph of her motion.→

Ans.

The women start from home to the office at 9.00 am from the origin O where t =9.00 am

The speed of the woman is given 5 km h-1 and the distance to the office is 2.5 km

Time taken to the office by the woman is = distance/speed = 2.5/5 = 0.5 h = 0.5×60 =30 minutes,therefore she will reach to the office at 9.30 am

She stays at office from 9.30 am to 5 pm,showing this period by a straight line parallel to x-axis since she is in stationary position within this period.

She return to her home at the speed of 25 km h-1

Time taken by her to reach home = distance of her home from the office/her speed = 2.5/25 = (1 /10)h  =6 min

She reach to her home from her office at (5 pm +6 min=5.06) pm.

Question 3. 4. A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.

Ans. The drunkard moves 5 steps forward and 3 steps backward

One step is equal to 1 m ,so he moves 5 m forward and 3 m backward in 8 s

The drunkard moves (5-3 =2 m) in 8 seconds

2 m is moved by the drunkard in 8 s

8 m will be travelled by him in 32 s

So in next 5 steps(i,e 5m) he will fall in the pit and the time taken to complete this journey is 32 +5=37s

Q 3. 5. A jet airplane travelling at the speed of 500 km/h ejects its products of combustion at the speed of 1500 km h-1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?

Ans. The velocity of the jet airplane is  Vjet,= 500 km/h

Velocity of the combustion products relative to Jet is Vcj ,= 1500 km/h

Let the speed of the combustion product with respect to observer is =x km/h

Velocity of combustion products with respect to jet = Velocity of combustion with respect to observer -Velocity of the jet with respect to the observer

-1500 = x – 500( negative sign of the combustion products is for opposite direction to the jet)

x = -1500 +500 =1000

Hence the velocity of the combustion products with respect to the observer is -1000 km/h,negative sign shows that its direction is opposite to the direction of the jet

Q3.6- A car moving along a straight highway with a speed of 126 km h–1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?

Ans. Initial speed of the car,u = 126 km h–1

The distance traveled by car,s = 200 m

Since the car is stopped after a distance, therefore its final speed,v =0

Applying the third equation of the motion

v² = u² + 2as

126 km/h =126000/3600 =35 m/s

0 = 35² + 2a(200)

400a = -1225

a = -1225/400 =-3.06 m/s²

v = u + at =35 – 3.06t

t = 35/3.06 =11.4 s

Therefore, the car stopped after 11.4 s with a retardation of -3.06m/s²

Q3. 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-1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 ms-1. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?

Ans.  The speed of both the train is given same, 71km/h therefore the relative velocity of train B with respect to the train A = UBA=0

The train B overtakes the train A after 50 s and the total distance covered by the train B is (s+800 m) where s is the initial distance between the train and 800 m is the sum of the lengths of both the train.

Applying second equation of the motion

s + 800 = UBAt + (1/2)at²

s +800 = 0 + (1/2)×1×50²

s = 1250 -800

s = 450 m

Therefore the original distance between guard of B from the driver of A is 450 +800 =1250 m

Q3.8-On a two-lane road, car A is travelling with a speed of 36 km h-1. Two cars B and C approach car A in opposite directions with a speed of 54 km h-1 each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?

Ans.The speed of the car A is ,=VA 36km/h =36000/3600 =10 m/s

The speed of the car B and C is same, VB=VC=54 km/h = 54000/3600 =15 m/s

The distance between the car B and A ,AB and the distance between the car A and C ,AC are 1 km =1000 m

Since the car B and the car C approaches the car A from opposite direction

The relative velocity of the car B with respect to the car A ,VBA= VB-VC = 15 -10 =5 m/s(i.e have same direction)

Relative velocity of car C w.r.t car A = 15 + 10 =25 m/s(since both are moving opposite to each other)

The time taken by the car C in overtaking the car A = Distance AC/Relative velocity of car C w.r.t car A =1000/25 =40s

In 40 seconds the car B has to overtake the car A,before the car C does, the car B accelerated by ‘a’ for doing this

Applying the second equation of the motion

S = uBAt + (1/2)at² = 5×40 + (1/2)×40²a

1000 =200 + 800a

a = 1

The minimum acceleration of car B is required to avoid an accident is 1 m/s²

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