Q77 / 90 · JEE Main 2023 (25 Jan Shift 2) · Physics

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

Solution

Given here, $x = 4 t^{2}$

We know that velocity is the rate at which displacement changes with time. So, $v = \frac{d x}{d t} = \frac{d (4 t^{2})}{d t} = 8 t$

Now, at $t = 5 s$

Velocity of particle is $v = 8 \times 5 = 40 m s^{- 1}$ .

Hence, the correct answer is option: 'A'.