JEE Physics 2023 Question with Solution

Given: The position is $x = 4t^2$.

Find: The velocity at $t = 5 \, \text{s}$.

Velocity is the derivative of position with respect to time.

$$v = \frac{dx}{dt} = \frac{d}{dt}(4t^2) = 8t$$

At $t = 5 \, \text{s}$,

$$v = 8 \times 5 = 40 \, \text{m/s}$$

Therefore, the computed velocity is $40 \, \text{m/s}$. The solution states the correct option is D, although this value matches option A; this is a discrepancy in the solution.

Given: $x = 4t^2$

Find: Velocity of the particle at $t = 5 \, \text{s}$.

Use the definition of instantaneous velocity:

$$v = \frac{dx}{dt}$$

Differentiate the given displacement function:

$$v = \frac{d}{dt}(4t^2)$$ $$v = 4 \cdot 2t = 8t$$

Now substitute $t = 5$:

$$v = 8(5) = 40 \, \text{m/s}$$

Hence, the physically correct value is $40 \, \text{m/s}$.