Asked by Anonymous
Find the distance, in feet, a particle travels in its first 4 seconds of travel, if it moves according to the velocity equation v(t)= −t2 + 4 (in feet/sec).
Answers
Answered by
Steve
s(t) = ∫ v(t) dt
= ∫ 4-t^2 dt
= 4t - 1/3 t^3 + C
Since s(0) = 0, C=0, and
s(t) = 4t - 1/3 t^3
= ∫ 4-t^2 dt
= 4t - 1/3 t^3 + C
Since s(0) = 0, C=0, and
s(t) = 4t - 1/3 t^3
Answered by
Stevesucks
Steve you honestly suck
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