Class 11 Physics

# Motion in Straight Line

## NCERT Solution

### Part 1

Question 1: In which of the following examples of motion, can the body be considered approximately a point object:

1. A railway carriage moving without jerks between two stations
2. A monkey sitting on top of a man cycling smoothly on a circular track
3. A spinning cricket ball that turns sharply on hitting the ground
4. A tumbling beaker that has slipped off the edge of a table.

Answer: (a) and (b) because in both cases size of the moving object is much smaller compared to path length

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 given figure. Choose the correct entries in the brackets below:

1. (A/B) lives closer to the school than (B/A)
2. (A/B) starts from the school earlier than (B/A)
3. (A/B) walks faster than (B/A)
4. A and B reach home at the (same/different) time
5. (A/B) overtakes (B/A) on the road (once/twice)

1. A lives closer to the school than B
2. A starts from the school earlier than B
3. B walks faster than A
4. A and B reach home at different time. (As orange and green lines show, they reach home at different times.)
5. B overtakes A on the road once

Question 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.

Question 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.

Answer: Drunkard takes 5 second in going 5 m and takes 2 second in coming back 3 m. Thus, he traverses 5 – 3 = 2 m in 8 second.

After 4 × 8 = 32 second, distance covered = 4 × 2 m = 8 m

Now, he needs to travel another 5 m to reach the pit. He will move 5 steps forward in another 5 second.

So, total time taken = 32 + 5 = 37 second

Question 5: A jet airplane travelling at the speed of 500 km h-1 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?

Answer: Relative speed of ejected product = 1500 km h-1

So, actual speed of ejected product = 1500 – 500 = 1000 km h-1

Question 6: A car moving along a straight highway with 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?

Answer: Given v0 = 126 km h-1 =126xx5/(18)=35ms^(-1)

Distance X = 200 m, v = 0, t = ? and a = ?

v^2=v_0^2+2aX

Or, 0=v_0^2+2aX

Or, X=-(v_0^2)/(2a)

Or, 200=-(35^2)/(2a)

Or, a = -(1225)/(2xx200)=-3.06 m s-2

Now, time can be calculated as follows:

v = v0 + at

Or, 0 = 35 – 3.06 t

Or, t=(35)/(3.06)=11.4 s