# Calculus

Find the distance traveled in 45 seconds by an object traveling at a velocity of v(t) = 20 + 2cos(t) feet per second.

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1. distance expression = 20t + 2sin(t) + c , where c is a constant
I will assume that you mean the distance in the first 45 seconds.
position at t=0
= 0 + 2sin0 + c = c
position at t = 45
= 900 + 2sin45 + c

distance travelled = 90+2sin45 + c - c
= appr. 901.7 feet

( notice in the second 45 seconds, the object would have travelled
1800 + 2sin90 + c - 900 - 2sin45 - x
= appr. 900.1 , different from the distance covered in the first 45 seconds.
If you wanted the actual distance of the path along the sine curve of the object you would have a much more difficult question)

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