JEE Physics 2024 Question with Solution

Given: Each car moves towards the other with initial speed $20 \, \text{m/s}$. The initial separation is $300 \, \text{m}$. Each car has retardation $2 \, \text{m/s}^2$.

Find: The distance between the cars when both come to rest.

Use the equation of motion for one car:

$$v^2 = u^2 + 2as$$

For one car, the final speed is $0$, initial speed is $20 \, \text{m/s}$, and acceleration is $-2 \, \text{m/s}^2$.

$$0 = (20)^2 + 2(-2)s$$ $$0 = 400 - 4s$$ $$s = \frac{400}{4} = 100 \, \text{m}$$

So each car travels $100 \, \text{m}$ before stopping. Therefore, the total distance covered by both cars together is

$$100 \, \text{m} + 100 \, \text{m} = 200 \, \text{m}$$

The remaining distance between them is

$$300 \, \text{m} - 200 \, \text{m} = 100 \, \text{m}$$

Therefore, the distance between the cars when they come to rest is $100 \, \text{m}$. The correct option is C.

Given: Let the distance travelled by each car before coming to rest be $d$. Each car has initial speed $20 \, \text{m/s}$ and deceleration $-2 \, \text{m/s}^2$.

Find: The distance between the cars when they stop.

Using the equation of motion:

$$v^2 = u^2 + 2ad$$

where $v = 0$, $u = 20 \, \text{m/s}$, and $a = -2 \, \text{m/s}^2$.

$$0 = 20^2 + 2(-2)d$$ $$0 = 400 - 4d$$ $$d = 100 \, \text{m}$$

Thus, each car covers $100 \, \text{m}$. Since the cars move towards each other, the total distance covered is

$$100 + 100 = 200 \, \text{m}$$

Hence the distance left between them is $300 - 200 = 100 \, \text{m}$. Therefore, the correct option is C.