Question

An automobile travelling with a uniform speed of 60 km/h on
applying the brakes comes to a stop at a distance of 20 m with uniform retardation. If the car doubles its initial speed, what will be the new stopping distance?​

Answer 1

Answer:

use

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

for both situation separately

Explanation:

then for the 1st situation

${v}^{2} = {u}^{2} + 2as \\ 0 = ({ \frac{50}{3} })^{2} - 2a20$

and for the 2nd situation

$0 = { \frac{100}{3} }^{2} - 2ax$

using the above 2 equation you can solve and get the answer

therefore the answer is 80m