A race car is slowed with a constant acceleration of 11 m/s2 opposite the direction of motion. a. If the car is going 55 m/s, how many meters will it travel before it stops? m b. How many meters will it take to stop a car going twice as fast? m; this is times farther than stopping a car going half the speed.

a. 137.5 m

b. 550 m

Explanation:

Parameters given:

Acceleration of car, a = - 11 m/s² (because it acts in opposite direction of the motion)

Initial velocity, u = 55 m/s

a. For the car to come to a stop, that means it's final velocity, v, will be 0 m/s. Using one of the equations of motion, we can find the distance traveled, s:

v² = u² + 2as

=> 0² = 55² + 2 * (-11) * s

=> 22*s = 55²

s = 55²/22

s = 137.5 m

b. If the car is moving at twice its initial velocity, u = 55 * 2 = 110 m/s:

v² = u² + 2as

0² = 110² + 2 * (-11) * s

=> 22s = 110²

s = 110²/22

s = 550 m