# 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:

- A railway carriage moving without jerks between two stations
- A monkey sitting on top of a man cycling smoothly on a circular track
- A spinning cricket ball that turns sharply on hitting the ground
- 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:

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

**Answer:**

- A lives closer to the school than B
- A starts from the school earlier than B
- B walks faster than A
- A and B reach home at different time. (As orange and green lines show, they reach home at different times.)
- 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.

**Answer:**

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 v_{0} = 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 = v_{0} + at

Or, 0 = 35 – 3.06 t

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