Posted by Andrea on Monday, December 19, 2011 at 10:29am.
a) The average speed during the interval is (1/2)(15+7) = 11 m/s
The time required to travel 90 m while decelerating from 15 to 7 m/s is
t = 90/11 = 8.18 s
The acceleration rate during that interval is
a = (7 - 15)/8.18 = -0.978 m/s^2
b) To stop, the additional time required is (7 m/s)/(0.978 m/s^2) = 7.158 s
Multiply that by the average speed during that interval, 3.5 m/s.
25.06 m/s
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