Question

Two boys A and B pass the same place P on a straight
road riding their bicycles in the same direction. The
boy A passes P at 7:30:00 AM and moves with
uniform velocity of 5 m/s. Another boy B passes
the place P some time after A, and moves with
uniform velocity 10 m/s and overtakes A, 500 m
away from P.
At what time does the boy B passes place P ?
(1) 7:30:30 AM (2) 7:30:40 AM
(3) 7:30:50 AM (4) 7:31:40 AM​

Answer 1

Answer:

The boy B will pass the position P at time T = 7 : 30: 50 AM

Explanation:

As we know that the distance moved by both A and B will 500 m when B overtakes A

Here we know that A travels with speed 5 m/s and it travels 500 m distance

so the time taken by A to reach the distance is given as

$t = \frac{d}{v}$

$t = \frac{500}{5}$

$t = 100 s$

now time taken by B to reach the same distance

$t' = \frac{500}{10}$

$t' = 50 s$

so time difference between two is given as

$\Delta t = 100 - 50$

$\Delta t = 50 s$

so B will start after 50 s so the time at which it will start is given as

T = 7 : 30 : 50 AM

#Learn

Topic : Distance and Motion

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