JEE Physics 2025 Question with Solution

Given: The distance is to be found from the velocity-time graph for the first $30.5 \, \text{s}$.

Find: Total distance covered in $30.5 \, \text{s}$.

The area under a velocity-time graph gives the distance travelled.

From the graph, the motion has two parts:

1. From $0$ to $2 \, \text{s}$, velocity increases linearly from $200 \, \text{m/s}$ to $400 \, \text{m/s}$.
2. From $2 \, \text{s}$ to $30.5 \, \text{s}$, velocity remains constant at $400 \, \text{m/s}$.

Distance in the first part:

$$d_1 = \left(\frac{200 + 400}{2}\right) \times 2 = 600 \, \text{m}$$

Distance in the second part:

$$d_2 = 400 \times (30.5 - 2) = 400 \times 28.5 = 11400 \, \text{m}$$

Total distance:

$$d = d_1 + d_2 = 600 + 11400 = 12000 \, \text{m}$$

Converting into kilometres:

$$d = \frac{12000}{1000} = 12 \, \text{km}$$

Therefore, the airplane covers $12 \, \text{km}$ in the first $30.5 \, \text{s}$. The correct option is D.

Given: A velocity-time graph with velocity changing from $200 \, \text{m/s}$ to $400 \, \text{m/s}$ in the first $2 \, \text{s}$, and then staying at $400 \, \text{m/s}$ up to $30.5 \, \text{s}$.

Find: Distance covered in the first $30.5 \, \text{s}$.

This graph can be treated as a trapezium from $0$ to $2 \, \text{s}$ and a rectangle from $2$ to $30.5 \, \text{s}$.

For $0$ to $2 \, \text{s}$:

$$d_1 = \left(\frac{v_i + v_f}{2}\right) t$$ $$d_1 = \left(\frac{200 + 400}{2}\right) \times 2 = 600 \, \text{m}$$

For $2$ to $30.5 \, \text{s}$:

$$d_2 = vt = 400 \times 28.5 = 11400 \, \text{m}$$

Hence,

$$d = 600 + 11400 = 12000 \, \text{m} = 12 \, \text{km}$$

The final answer is $12 \, \text{km}$, so the correct option is D.

The other solution path on the page gives $11.8 \, \text{km}$ by incorrectly taking the first segment as a triangle from zero velocity. The graph actually starts at $200 \, \text{m/s}$, so the correct first area is a trapezium, not a triangle.