2016 01 Level 1 or 2?

Classical Mechanics Level 3

A bus starts from rest moves with an acceleration 1 m/ s 2 1 \text{ m/ s}^{ 2 } . A man who is 48 m 48\text{m} behind the bus is running with 10 m/s 10\text{ m/s} . Calculate the minimum time (in seconds) taken by him to catch the bus.

Bonus Can you explain why there are two time instances where he can board the bus?

More questions??


The answer is 8.

Anandhu Raj by Anandhu Raj , 23, India

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2 solutions

Anandhu Raj
Feb 2, 2016

Let x x be the distance traveled by bus before the man board on it.Let t be the time taken for it.

Displacement of man = 48 + x = 10 t = 48+x=10t ------ (1)

Displacement of bus = x = 1 2 × 1 × t 2 = x= \frac { 1 }{ 2 } \times 1\times { t }^{ 2 } ------- (2)

Using (1) and (2) ,

t=8s or t=12s

There are two time instances because at first the man can overtakes the bus as his velocity is more initially.He could board the bus without overtaking.This corresponds to 8 t h { 8 }^{ th } second. Then the bus accelerates and its can overtake the man.The man could also board the bus at this instance also. This corresponds to 12 t h { 12 }^{ th } second.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice explanation... Another method would be to use Relative Motion.

Rishabh Jain - 5 years, 4 months ago
Ramiel To-ong
Feb 3, 2016

nice problem

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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