Question

1. Now suppose the car doubles its speed to 25.0 m/s (~90 km/hr). Its brakes can apply a force of 5.00x103 N.  What is the minimum stopping distance at this speed?

Answer 1

Apply Newton's second law to determine deceleration:

$$a=\frac Fm.$$

The minimum stopping distance at this speed is

$$d=\frac{v^2}{2a}=\frac{v^2m}{2F}=0.125m,$$

where m is the mass of the car.