Motion In a Straight Line - 11th physics

Question (3.1) 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.

Solution

Here the option (a), (b) and (c) the given objects in motion are considered approximately a point object.

Explanation

While considering the motion, the objects in motion are treated as point objects. This approximation is valid so far as the size of the object is much smaller than the distance it moves in a reasonable duration of time.

This means the object in motion is conventionally considered as point object compared to the distance it goes.

In option (a) since the distance between two stations is very large compared to the railway carriage moving between distance of two stations. Thus the railway carriage can be considered as a point object.

In option (b) a monkey sitting on the top of a man cycling on a circular track. The monkey can be considered as point object.

In option (c) a spinning cricket ball that turns sharply on hitting the ground is considered as point object compared to the distance it moves through the ground.

In option (d) a tumbling beaker that has slipped off the edge of a table is not considered as point object as size of beaker is not very small compared to the distance it moves.

Question (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 Figure. Choose the correct entries in the brackets given below:

(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

(a) A lives closer to school than B

Explanation From given position-time graph it is clear that OP is less than OQ. This means that A travels less distance than B to reach home from school.

That means distance of home of A is less than the distance of B from schoo.

And hence, A lives closer to school than B.

(b) A starts earlier than B from schoo.

Explanation From given position-time graph it is clear that A starts from x=0 and t=0. While B starts from school at x=0 and t=some time (some value after t=0).

Thus, A starts earlier than B from school.

(c) B walks faster than that of A

Explanation In position-time graph the slope of graph shows the speed. Steeper the slope faster is the speed and vice versa.

From given position-time graph it is clear that slope of B is steeper than the slope of A.

Thus, B travels faster than the A

(d) B overtakes A once on the road

Explanation From given position-time graph it is clear that the slope of graph of B is steeper than A. This means B walks faster than that of B. And since graphs meet once only.

Thus, B overtakes A once only on the road.

Question (3.3) A woman starts from her home at 9.00 am, walks with a speed of 5 kmh^–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 kmh^–1. Choose suitable scales and plot the x-t graph of her motion.

Solution

Given, time to start by women from home = 9.00 am

Speed of women while going to office = 5 km/h

Distance of office = 2.5 km

Time upto stay in office = 5.00 pm

Return back time from office = 5.00 pm

Speed while returning from office = 25 km/h

We know that, Time taken = Distance/Speed

Thus, time taken to reach the office

`=(2.5\ km)/(5\ km//h)`

`=1/2h=30m`

Thus, reaching time at office = Starting time + Time taken

= 9.00 am + 30 m

Thus, reaching time at office = 9.30 am

And time taken to return from office

`=(2.5\ km)/(25\ km//h)`

`=1/10h=6m`

Thus, time to come back from office = starting time from office + Time taken

= 5.00 pm + 6 m

Thus, time to come back from office = 5.06 pm

Thus, using above data position-time (x-t) graph of women can be drawn as below.

Question (3.4) A drunkard waking 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 1s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13m away from the start.

Solution

Given, Drunkard takes 5 steps forward and 3 steps backward

Length of each step = 1m

Time to take 1 step = 1s

Distance of pit in the way = 13m

Thus, time to fall in pit = ?

In one move distance covered = Length of forward steps – Length of backward steps

= 5 m – 3 m

Thus, distance covered in first forward and backward move = 2m

Time taken to move first 5 steps forward = 5s

And time taken to return first 3 steps backward = 3s

Thus, total time taken to cover first 2 m = 8s

Similarly, time taken to cover second 2 m i.e. total 4m = 8s × 2 = 16s

Similarly, time taken to cover next 8m = 16s × 2 = 32s

Now, as given after taking backward steps drunkard takes 5 steps forward

Now, as given the distance of pit = 13m

Thus, distance left = 13m – total distance covered in 32s

= 13m – 8m = 5m

Since, as given in question drunkard takes 5 steps forward and each step is 1m long and he takes each step in 1s

Thus, time taken to reach the last 5m = 5s

Thus, total time taken to fall in pit = time taken to cover first 8m + time taken to cover last 5m

= 32s + 5s = 37s

Thus, after 37s from start the drunkard falls in the pit.

The x-t graph of drunkard

Question (3.5) 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?

Solution

Given, speed of jet airplane (v_jet) = 500 km/h

Speed of products of combustion (v_pc) = 1500 km/h

Thus, relative speed of product of combustion to the observer on the ground (v_o) = ?

Since velocity of the product of combustion (v_pc) is in opposite direction of plane, thus it has a negative sign.

i.e. v_pc = –1500 km/h

Now,

Speed of product of combustion to the observer on the ground (v_o) = Speed of product of combustion (v_pc) + speed of jet plane (v_jet)

⇒ v_o = –1500 km/h + 500 km/h

⇒ v_o = –1000 km/h

Here, negative sign of the speed of product of combustion shows that it is going in the opposite direction to the jet plane.

Thus, speed of product of combustion respect to the observer on the ground = –1000 km/h Answer

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