Kinematics

JEE Main level, single correct, medium

126 of 308

Question 126 of 308

The distance travelled by a particle is related to time tt as x=4t2x=4{t}^{2}. The velocity of the particle at t=5st=5s is

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Answer: (A) 40ms–140m{s}^{–1}

Given here, x=4t2x=4{t}^{2}
We know that velocity is the rate at which displacement changes with time. So, v=dxdt=d(4t2)dt=8tv=\dfrac{dx}{dt}=\dfrac{d\left(4{t}^{2}\right)}{dt}=8t
Now, at t=5st=5s
Velocity of particle is v=8×5=40ms−1v=8\times 5=40m{s}^{-1}.

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