Find the distance traveled in 35 seconds by an object traveling at a velocity of v(t) = 20 + 4cos(t) feet per
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Find the distance traveled in 35 seconds by an object traveling at a velocity of v(t) = 20 + 4cos(t) feet per

[From: ] [author: ] [Date: 11-09-08] [Hit: ]
Note that in this case the absolute value really wasnt necessary because cos(t) can only range from [-1,You can use the TI-83 or the TI-89 definite integration features to obtain an answer.-I didnt do this integral by hand... I plugged it into my calculator (TI-89).......
Consider the following problem.

Find the distance traveled in 35 seconds by an object traveling at a velocity of v(t) = 20 + 4cos(t) feet per second.

I know the answer is 700 but I don't know How or Why...

Please Help.

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Remember that when you want to calculate distance, you want to take the integral of the ABSOLUTE VALUE of the velocity function to make sure that all that area underneath the x-axis is reflected about the x-axis so that the area becomes positive. In this problem:

Distance = ∫|v(t)| dt from a to b

= ∫|20 + 4*cos(t)| dt from 0 to 35 = 698.29 feet or about 700 feet

Note that in this case the absolute value really wasn't necessary because cos(t) can only range from [-1,1] and 4 multiplied by a number in that range is < 20.

You can use the TI-83 or the TI-89 definite integration features to obtain an answer.

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I didn't do this integral by hand... I plugged it into my calculator (TI-89).

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