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October 30, 2014

October 30, 2014

Posted by **Tri** on Monday, September 10, 2007 at 3:41am.

- Physics -
**drwls**, Monday, September 10, 2007 at 7:50amThe distance travelled while accelerating at a rate a=14 m/s^2 is X = 450 m and the maximum velocity attained at that time is V = sqrt(2aX) = 112.2 m/s. The time required to decelerate to a stop at a rate a' = 7 m/s^2 is t = V/a' = 16.0 s

The average velocity while decelerating is

v = V/2 = 56.1 m/s and the distance travelled while declerating is v*t

X' = 56.1 m/s*16 s = 900 m

The total distance travelled when it has stopped is 450 + 900 = 1350 m

Here is an easier way: Since the rate of deceleration is half the rate of acceleration, it takes twice as long to decelerate and one travels twice as far (900 m)while decereating.

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