A bus at rest starts accelerating at 1 m/s. A man 50 m behind it runs to catch the bus. At what minimum speed should he run to catch the bus

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Solution

at the time when the passenger will catch the bus velocity must also be V.
so the final velocity of the bus is V and initial velocity is 0
so from eqn of motion we have
V=0+at
t=V/a
distance covered by the bus in t time
V^2-0=2aS
S=V^2/2a
so the the passenger have to run d+S distance in time t.
then minimum velocity will be
V=d+S/t
V=(d+V^2/2a)/t
V=(d+V^2/2a)/(V/a)
put a=1m/s^2
d=50m
V=(50+V^2/2)/V
V^2=50+V^2/2
2V^2=100+V^2
2V^2-V^2=100
V^2=100
V=10m/s

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