Posted by **Priscilla** on Sunday, December 27, 2009 at 7:29pm.

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?

- Physics -
**Priscilla**, Sunday, December 27, 2009 at 8:34pm
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?

- Physics -
**drwls**, Sunday, December 27, 2009 at 10:55pm
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.

- Physics -
**josh**, Tuesday, April 9, 2013 at 8:52am
56 mph7 34mph

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