A woman driving a 2.0 x 10^3 kg car at 15m/s fully applies the brakes 50.m from a stop light. If the car stops 5.0m before the light, what is the magnitude of the average force applied by the brakes? (4.5 x 10^3)??

Question

A woman driving a 2.0 x 10^3 kg car at 15m/s fully applies the brakes 50.m from a stop light.  If the car stops 5.0m before the light, what is the magnitude of the average force applied by the brakes?    (4.5 x 10^3)??

stormfire1 · Answer

$$Vi=15 m/s$$$$Vf = 0m/s$$$$Xf-Xi=0m-45m=-45m$$Use the following kinematic equation to find the (negative) acceleration caused by her braking:
$$V_f^2-V_i^2=2a(X_f-X_i)$$Solve for a to get:$$a=\frac{(-15m/s)^2}{-90m}=-2.5 m/s^2$$Now find the force:$$F=ma=(2000kg)(-2.5 m/s^2)=-5000N$$
The magnitude of the force applied by her brakes is therefore 5000N.