A 1000-kg car experiences a net force of 9500 N while slowing down from 30 m/s to $15.9 \mathrm {~m} / \mathrm { s }$ How far does it travel while slowing down?
A)34 m
B)31 m
C)37 m
D)41 m

Question

Quizplus · Accepted Answer

The work done by the net force on the car is equal to the change in the car's kinetic energy:$$W = Fd = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$where:* W is the work done by the net force* F is the net force* d is the distance traveled* m is the mass of the car* v_i is the initial velocity of the car* v_f is the final velocity of the carSolving for d, we get:$$d = \frac{1}{2}\frac{mv_f^2 - mv_i^2}{F}$$Plugging in the given values, we get:$$d = \frac{1}{2}\frac{(1000 	ext{ kg})(15.9 	ext{ m/s})^2 - (1000 	ext{ kg})(30 	ext{ m/s})^2}{9500 	ext{ N}} = 34 	ext{ m}$$Therefore, the car travels 34 m while slowing down.

A 1000-Kg Car Experiences a Net Force of 9500 N 15.9 m/s15.9 \mathrm {~m} / \mathrm { s }15.9 m/s

A 1000-Kg Car Experiences a Net Force of 9500 N $15.9 \mathrm {~m} / \mathrm { s }$