two friends started from same point with some constant acceleration of 5m/s second one starts journey 5 sec later. find the time taken by 1 St one such that he is 100m ahead of 2 nd one
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Answer: 6.5 s
Explanation: If the time traveled upto the given point by the second one is seconds, the first one has takent+5seconds.
Since the difference between their distances travelled is 100m,
\dfrac{1}{2}a(t+5)^2-\dfrac{1}{2}at^2=100
\implies \dfrac{1}{2}a(10t+25)=100
\implies t=1.5s
Hence time taken by first one is6.5s
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