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A bus starts to move with acceleration 1m/square. A man who is 48m behind the bus runs to catch it with constant velocity 10m/s.In how much time he will catch the bus.
Asked by arunkumarsingh3055 | 30 Jan, 2019, 10:43: PM
answered-by-expert Expert Answer
Let the man catches the bus after t seconds. Distance travelled by the man before catching bus = 10t m
 
Distance travelled by the bus in t seconds = (1/2)a×t2 , where a is acceleration = 1 m/s2

since at starting time, man is 48 m behind the bus,  we  have  (1/2)×t2 + 48 = 10t ,   or  t2 -20t +96 = 0 ...........(1)

by factorising LHS of eqn.(1), we write,  (t-8)(t-12) = 0
 
hence t = 8, after 8 seconds man will catch the bus
Answered by Thiyagarajan K | 30 Jan, 2019, 11:23: PM

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