Question

The velocity function, in feet per second, is given for a particle moving along a straight line.
v(t) = t2 − t − 132, 1 ≤ t ≤ 15
Find the total distance that the particle travels over the given interval

Answer 1

Integrate v(t), but when you solve v(t) = 0 and you'll see there's a root at t = 12, so break up the integral.

\[\left| \int\limits_{1}^{12}v(t)~dt \right|+\left| \int\limits_{12}^{15}v(t)~dt \right|\]

Answer 2

I got 3084.667 why is the not right

Answer 3

I don't know what you did. can you put up your work?

Answer 4

\[\int\limits \frac{ t^{3} }{ 3 }-\frac{ t^{2} }{ 2 }+132t\]

thats what I used for my integral then I plugged in 1 & 12, then subtracted the 1 value from the 12 value. Then I plugged in 15 and subtracted the 12 value from that & then added the two subtractions

Answer 5

\[\left| \frac{ t^3 }{ 3 }-\frac{ t^2 }{ 2 }-132t \right|_1^{12}=\left| -1080-(\frac{ -793 }{ 6 }) \right|=\frac{ 5687 }{ 6 }\]

\[\left| \frac{ t^3 }{ 3 }-\frac{ t^2 }{ 2 }-132t \right|_{12}^{15}=\left| -\frac{ 1935 }{ 2 }-(-1080) \right|=\frac{ 225 }{ 2 }\]

Answer 6

add them together to get 381/3 ≈ 1060.3 ft

Answer 7

but 12 is the zero so you would be adding that integral

Answer 8

yes, you want to add the two area to get the total distance traveled