A car is traveling at upsilonx 24 ms The driver applies the

A car is traveling at upsilon_x = 24 m/s. The driver applies the brakes and the car decelerates at a_z = -4.0 m/s^2.What is the stopping distance? 3.0 m 97 m 62 m 72 m

Solution

Using the equation of motion,

v^2 = u^2 + 2as

here

s = stopping distance

u = 24 m/s

a = -4 m/s2

v = 0

So, 0^2 = 24^2 +2*(-4)*s

So, s = 72 m <-----answer