Two cars are travelling towards each other at speed of $20 \mathrm{~m} \mathrm{~s}^{-1}$ each. When the cars are $300 \mathrm{~m}$ apart, both the drivers apply brakes and the cars retard at the rate of $2 \mathrm{~m} \mathrm{~s}^{-2}$. The distance between them when they come to rest is :

<p>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 $2 \, \text{m s}^{-2}.$</p> <p>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:</p> <p>$v_f^2 = v_i^2 + 2ad$</p> <p>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):</p> <p>$0 = v_i^2 + 2ad \Rightarrow d = -\frac{v_i^2}{2a}$</p> <p>Plugging in the values for each car (noting that acceleration $a$ is negative because it is deceleration, so $a = -2 \, \text{m s}^{-2}$):</p> <p>$d = -\frac{(20)^2}{2(-2)} = -\frac{400}{-4} = 100 \, \text{m}$</p> <p>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 $2 \times 100 \, \text{m} = 200 \, \text{m}$.</p> <p>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:</p> <p>$300 \, \text{m} - 200 \, \text{m} = 100 \, \text{m}$</p> <p>So, the distance between the cars when they come to rest is $100 \, \text{m}$. Therefore, the correct option is:</p> <p>Option B 100 m</p>