Question

A car going 50 ms is brought to rest in a distance of 20.0 m as it strikes a pile of dirt. How large

an average force is exerted by seatbelts on a 90 kg passenger as the car is stopped?

Answer 1

Let's first find the deceleration of the car:

$$v^2=v_0^2+2as,$$ $$a=\dfrac{v^2-v_0^2}{2s}=\dfrac{0-(50\ \dfrac{m}{s})^2}{2\cdot20\ m}=-62.5\ \dfrac{m}{s^2}.$$

The sign minus means that the car decelerates.

Finally, we can find an average force exerted by seatbelts on a 90 kg passenger as the car is stopped:

$$F=ma=90\ kg\cdot(-62.5\ \dfrac{m}{s^2})=-5625\ N.$$

The magnitude of the average force exerted by seatbelts on a 90 kg passenger as the car is stopped equals 5625 N. The sign minus means that the average force directed in the opposite direction to the motion of the car.