Question

A bus starts from rest with an acceleration of 1ms−2 A man who is 48m behind the bus starts with a uniform velocity of 10ms−1. The minimum time after which the man will catch the bus is

Answer 1

To find the time taken by using Newton's second equation of motion.
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Acceleration (a) = 1m/s^2

Distance(s) = 48 m

Initial velocity (u) = 10m/sec

time taken =?
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By using Newton's second equation of motion.

S= ut + 1/2a(t)^2

48= 10×t + 1/2×1×(t)^2

48= {2×10t + (t)^2}/2

96= 20t + (t)^2

(t)^2+20t-96 = 0

(t)^2 +(16+6)t-96 = 0

(t)^2 + 16t + 6t -96 = 0

t (t+16) + 6 (t -16)=0

(t+16) + (t-6) = 0

So here cannot be negative. We use

t-6 = 0

t = 6 sec will be time taken.