Question

A bus starting from rest with an acceleration of 1m/sec^2. a man who is 48m behind

Answer 1

ANS - 8 seconds

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.

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Answer 2

Hy ...
Friend...

Thankyou you asking question...
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Here your answer
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Answer
relative velocity of man with respect to bus is V=vman-v bus=10-0=10m/s
relative accleration of man ......................is A=aman-abus=0-1= -1m/s^2.

relative displacement of man .................is 48 metres.

now using eq of motion for this relative motion .

s=ut+1/2at^2

we get t=8,12

minimum time will bw 8 sec.

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