Asked by Ella
A particle moves on a line so that its position at time t seconds is p(t)=3t3 meters to the right of the origin. Find the time c in the interval (4,8) for which the velocity of the particle at time c is equal to the average velocity on the interval [4,8].
Answers
Answered by
Steve
p(8) = 1536
p(4) = 192
so, avg v is (1536-192)/(8-4) = 336
p'(t) = 9t^2
so, you want
9c^2 = 336
p(4) = 192
so, avg v is (1536-192)/(8-4) = 336
p'(t) = 9t^2
so, you want
9c^2 = 336
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