Physics
posted 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?

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? 
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. 
56 mph7 34mph
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