JEE Physics 2024 Question with Solution

Given: $S = 2t^2 \hat{j} + 5t \hat{k}$ and $t = 1 \, \text{s}$.

Find: The velocity at $t = 1 \, \text{s}$ and the matching option.

Velocity is obtained by differentiating position with respect to time:

$$\vec{v} = \frac{d\vec{S}}{dt} = \frac{d}{dt}(2t^2 \hat{j} + 5t \hat{k})$$

So,

$$\vec{v} = 4t \hat{j} + 5 \hat{k}$$

At $t = 1 \, \text{s}$,

$$\vec{v} = 4 \hat{j} + 5 \hat{k}$$

Thus, the velocity has a $y$-component of $4 \, \text{m/s}$. From the given options, this matches $4 \, \text{m/s}$ in $y$-direction.

The solution also indicates a discrepancy because the full magnitude from

$$|\vec{v}| = \sqrt{4^2 + 5^2} = \sqrt{41}$$

would not equal any listed option. Therefore, based on the extracted working and option matching, the correct option is D.

Given: Position vector of the ant in the $Y-Z$ plane is $S = 2t^2 \hat{j} + 5t \hat{k}$.

Find: Velocity at $t = 1 \, \text{s}$.

Differentiate each component separately:

$$\frac{d}{dt}(2t^2 \hat{j}) = 4t \hat{j}$$ $$\frac{d}{dt}(5t \hat{k}) = 5 \hat{k}$$

Hence,

$$\vec{v} = 4t \hat{j} + 5 \hat{k}$$

Substituting $t = 1$,

$$\vec{v} = 4 \hat{j} + 5 \hat{k}$$

So the ant moves with a $y$-component $4 \, \text{m/s}$ and a $z$-component $5 \, \text{m/s}$.

Among the listed options, the matching statement is $4 \, \text{m/s}$ in $y$-direction.

Therefore, the correct option is D.