Question

I will make the Brianliest answer. please explain with steps . A man is behind a bus which is at rest The bus accelerating of the rate of 1m/s²at the the same time the man starts running at uniform velocity of 10m/s What is the minimum time is which the man catches the bus

Answer 1

Let's assume that the man catches the bus in t sec

Distance travelled by the bus in this time period
=[1/2]at²
=0.5x1xt²
=0.5t²

Distance travelled by the man in this time period
=10xt

Now both the distance must be same
so,

0.5t²=10t
t=20s

So, the man will catch the bus in 20 seconds.

Answer 2

Hey mate!

Here's your answer!!

In BUS FRAME-

Velocity of man = 10 m/s

s = 48 m

Acceleration of man with respect to bus = Acceleration of man - Acceleration of bus = 0 - (1) = -1 m/s

Applying second equation of motion,

s = ut + 1/2 at2

48 = 10t - 1/2 × t2

Solving we get, t = 12 sec or t = 8 sec

Therefore, the minimum time is 8 seconds.

$hope \: it \: helps \: you...$
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