JEE Physics 2025 Question with Solution

Given: $\mathbf{r} = \left( 5t^2 \hat{i} - 5t \hat{j} \right) \, \text{m}$

Find: The magnitude and direction of velocity at $t = 2 \, \text{s}$.

Velocity is the time derivative of the position vector.

$$\mathbf{v} = \frac{d}{dt}\left( 5t^2 \hat{i} - 5t \hat{j} \right) = 10t \hat{i} - 5 \hat{j}$$

At $t = 2 \, \text{s}$,

$$\mathbf{v} = 20 \hat{i} - 5 \hat{j} \, \text{m/s}$$

Now, the magnitude is

$$|\mathbf{v}| = \sqrt{20^2 + (-5)^2} = \sqrt{400 + 25} = \sqrt{425} = 5\sqrt{17} \, \text{m/s}$$

For direction with the $-\hat{y}$ axis,

$$\tan \theta = \frac{|v_x|}{|v_y|} = \frac{20}{5} = 4$$

So,

$$\theta = \tan^{-1}(4)$$

the solution gives this direction from the $-\hat{y}$ axis. However, the listed options contain $\tan^{-1}\left(\frac{5}{4}\right)$ instead of $\tan^{-1}(4)$. The magnitude matches option A, and the option is also marked correct in the source solution.

Therefore, the correct option is A.

Given: $\vec{r} = ( 5t^2 \hat{i} - 5t \hat{j}) \, \text{m}$

Find: The magnitude and direction of the velocity at $t = 2 \, \text{s}$.

1. Differentiate the position vector with respect to time:

$$\vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt} \left( 5t^2 \hat{i} - 5t \hat{j} \right)$$ $$\vec{v} = 10t \hat{i} - 5 \hat{j}$$

2. Substitute $t = 2 \, \text{s}$:

$$\vec{v}(2) = 10(2) \hat{i} - 5 \hat{j} = 20 \hat{i} - 5 \hat{j} \, \text{m/s}$$

3. Compute the magnitude:

$$|\vec{v}| = \sqrt{(v_x)^2 + (v_y)^2}$$ $$|\vec{v}| = \sqrt{(20)^2 + (-5)^2} = \sqrt{400 + 25} = \sqrt{425} = 5\sqrt{17} \, \text{m/s}$$

4. Find the angle with the negative $\hat{y}$-axis:

$$\tan \theta = \frac{|v_x|}{|v_y|} = \frac{|20|}{|-5|} = \frac{20}{5} = 4$$ $$\theta = \tan^{-1}(4)$$

Hence the velocity has magnitude $5\sqrt{17} \, \text{m/s}$ and is directed at an angle $\tan^{-1}(4)$ with the $-\hat{y}$ axis. Since the source marks A as correct despite the angle expression mismatch in the options, we select A.