Speeds of two identical cars are $U$ and $4 U$ at a specific instant. Assuming same retardation, the ratio of the respective distances in which the two cars are stopped from that instant is:

Using the third equation of motion, v^2 equals u^2 minus 2as, where the final velocity v is 0, the initial velocity is u, and the retardation is a. Rearranging for stopping distance s, we get s equals u^2 divided by 2a. For the two cars with initial speeds U and 4U and the same retardation, the ratio of their distances is U^2 divided by (4U)^2. Simplifying this gives 1 divided by 16, making the ratio 1:16.

• Thanks
• Wrong explanation