Question

A car traveling at 91.0 km/h approaches the turnoff for a restaurant 30.0 m ahead. If the driver slams on the brakes with an acceleration of -6.40 m/s^2, what will be her stopping distance?

Answer 1

We can find the stopping distance of the car from the kinematic equation:

$$v^2=v_0^2+2ad,$$ $$0=v_0^2+2ad,$$ $$d=\dfrac{-v_0^2}{2a}=\dfrac{-(25.28\ \dfrac{m}{s})^2}{2\times(-6.4\ \dfrac{m}{s^2})}=50\ m.$$