Kinematics

JEE Main level, single correct, medium

17 of 308

Question 17 of 308

The position of a particle related to time is given by x=(5t2−4t+5)mx=(5{t}^{2}-4t+5)m. The magnitude of velocity of the particle at t=2t=2 s will be :

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Answer: (D) 16ms−116m{s}^{-1}

It is given that
x=5t2−4t+5x=5{t}^{2}-4t+5
The velocity is given by

v=dxdt=d(5t2−4t+5)dt=10t−4v=\dfrac{dx}{dt}=\dfrac{d(5{t}^{2}-4t+5)}{dt}=10t-4
At
t=2st=2s,
v=10(2)−4=16ms−1v=10(2)-4=16m{s}^{-1}

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