a man is 48m behind a bus which is at rest. the bus starts accelerating at the rate of 1m/s2; at the same time the man starts running with uniform velocity of 10m/s. what is the minimum time in which the man catches the bus
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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
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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