QJEEM24Phe1d7: JEE Main 2024 Physics - Class 11 JEEM24Phe1d7 Solution: Two cars are travelling towards each other at speed of each.…

Let's analyze the given information before determining the distance between the two cars when they come to rest. Each car is traveling towards the other at a speed of $20 \, \text{m s}^{-1}$ and they both start braking when they are $300 \, \text{m}$ apart. The deceleration (negative acceleration) of each car is given as $2 \, \text{m s}^{-2}.$

To find the distance each car travels before coming to rest, we can use the kinematic equation that relates initial velocity ( $v_i$ ), final velocity ( $v_f$ ), acceleration ( $a$ ), and distance ( $d$ ), which is:

$v_f^2 = v_i^2 + 2ad$

Since the final velocity $v_f = 0$ (they come to rest), we can rearrange the equation to solve for $d$ (the distance each car travels before stopping):

$0 = v_i^2 + 2ad \Rightarrow d = -\frac{v_i^2}{2a}$

Plugging in the values for each car (noting that acceleration $a$ is negative because it is deceleration, so $a = -2 \, \text{m s}^{-2}$ ):

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

Each car travels $100 \, \text{m}$ before coming to rest. Since they both start $300 \, \text{m}$ apart and each travels $100 \, \text{m}$ towards the other, the total distance covered by both cars before stopping is $2 \times 100 \, \text{m} = 200 \, \text{m}$ .

To find the distance between them when they come to rest, we subtract the total distance covered by both cars from the original distance between them:

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

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

Option B 100 m