Question

A bus is begins to move with an acceleration of 1 m/s2. A boy who is 42m behind the bus starts running at 10 m/s towards the bus. After what time bus will cross the boy?

Answer 1

velocity of man = 10 m/s
s = 42 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
42= 10t - 1/2*t2
42= (20t-1t^2)/2
84= 20t - 1t^2
1t^2 - 20t +84 = 0
It is in the form of quadratic equation..
Therefore t=14 and t=6
We need minimum value
So take 6

Answer 2

hello friend....!!!

• According to the question,
we should calculate after what time bus will cross the boy.

• Let us consider, the distance covered by the bus, after 't ' seconds is,

1/2t^2

so therefore the total distance that should be covered is 42 + 1/2t^2

we know,

speed = distance / time

distance = speed X time

D = S X T

implies,

10t = 42 + t^2 / 2

by forming a quadratic equation we get,

20t = 84 + t^2

t^2 -20t + 84 = 0

we get t = 14 seconds and t = 6 seconds.


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hope it helps...!!!