Kinematics

JEE Main level, single correct, medium

18 of 308

Question 18 of 308

The distance travelled by an object in time tt is given by s=(2.5)t2s=\left(2.5\right){t}^{2} . The instantaneous speed of the object att=5st=5s will be :

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Answer: (A) 25ms–125m{s}^{–1}

It is given that
s=2.5t2s=2.5{t}^{2}
Differentiating,

v=dsdt=d(2.5t2)dt=5tv=\dfrac{ds}{dt}=\dfrac{d\left(2.5{t}^{2}\right)}{dt}=5t
Att=5s,v=25ms−1Att=5s,v=25m{s}^{-1}

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