Kinematics

JEE Main level, single correct, medium

47 of 308

Question 47 of 308

A particle starts with an initial velocity of 10.0ms−110.0{ms}^{-1} along xx-direction and accelerates uniformly at the rate of 2.0ms−22.0m{s}^{-2}. The time taken by the particle to reach the velocity of 60.0ms−160.0m{s}^{-1} is _____.

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Answer: (A) 25s25s

The given data is
v=60ms−1u=10ms−1a=2ms−2v=60m{s}^{-1} u=10m{s}^{-1} a=2m{s}^{-2}
Using the equation v=u+atv=u+at, we get
60=10+2t⇒t=(60−10)2⇒t=25s60=10+2t \Rightarrow t=\dfrac{\left(60-10\right)}{2} \Rightarrow t=25s

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