Get an expression for stopping distance of a vehicle in terms of intial velocity `v_(a)` and deceleration "a"…

Explanation

Let s be the distance travelled by a vehicle before it stops
Using `v^(2)-u^(2)=2as,`
`0_(0)^(2)=-2as`
`s=(v_(0)^(2))/(2a)`
The stopping distance is directily proportional to `v_(0)^(2)`. By doubling intial velocity it increase stopping distance by 4 times. Provided deceleration is kept as constant.