Asked by Priscilla


A race car can be slowed with a constant acceleration of -11 m/s^2

a. If the car is going 55 m/s, how many meters will it travel before it stops?

b. How many meters will it take to stop a car going twice as fast?

I'm confused a little, for these two problems they both want distance? or just time for a?
Answered by Priscilla
ok like one of the teachers previously stated to find the time.

V = Vi + at
0 = 55 - 11t
55 = 11t
5 = t

But for this problem i think you have to find two distance.

A, to find the distance
B, to find the distance i guess if the time is doubled, right?
Answered by drwls
A. stopping distance = (stopping time) x (average speed)
= (V/a)*(V/2) = V^2/(2a)

B. If the speed doubles, with deceleration rate "a" staying the same, the stopping distance is four times farther.
Answered by josh
56 mph7 34mph