Question

A bus starts from rest with an acceleration of 1m /sec square.a man who is 48m behind the bus starts with a uniform velocity of 10 m / sec,then the minimum time after which the man will catch the bus is...
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Answer 1

Hi friend,

# A man can catch the bus in 8 sec

➡Solution :-

✔Velocity of the man = 10m/s

S = 48m

✔Formula :- Acceleration of man with respect of bus = Acceleration of man - Acceleration of bus = 0-(1) = -1m/s


So, here we applying , second equation of motion,

S = ut + 1/2 at^2

➡48 = 10t - 1/2*t^2

➡Or, 10t = 48 + 1/2 + 1 × t^2

➡Or, t^2 - 20t + 96 =0

➡Or, t^2 - 8t- 12t + 96 =0

➡Or, t ( t-8) - 12(t-8) = 0

➡Or, (t-8) ( t-12) = 0

➡Or , t = 8 and 12s ( proved)

But we are interested in minimum time so, the man catch the in 8s minimum time...i hope it helps you
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Answer 2

Answer:

I hope it is helpful

please mark as brainlist plz