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3 years ago

Find the distance traveled in 35 seconds by an object traveling at a velocity of v(t) = 20 + 5cos(t) feet per second

1 answer:

8 0

$s(t)= \int\limits^{35}_0 {v(t)} \, dt \\ =\int\limits^{35}_0 {(20+5cos(t))} \, dt \\ =[20t+5sin(t)]^{35}_0 \\ =(20(35)+5sin(35))-(20(0)+5sin(0)) \\ =700+2.868=702.868 \, feet$

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