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The velocity function, in feet per second, is given for a particle moving along a straight line. v(t) = t^3 − 9t^2 + 20t − 12, 1 ≤ t ≤ 7 Find the total distance that the particle travels over the given interval.
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Distance is integral of velocity because velocity is the derivative of distance over time: $$ d=\int_1^7v(t)dt=\int_1^7\left (t^3 − 9t^2 + 20t − 12\right )~dt $$ Can you evaluate this integral?
yesssss I did I don't know why my answer is wrong
that will give you displacement think about something like: \(\int_{1}^{7}|v(t)|dt\)
|dw:1444090723561:dw|
yes then it changes at (t-1) (t-2) (t-6)
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it's in the interval \(1 ≤ t ≤ 7\) so the \(2 \le t \le 6\) is the bit that needs to be thought about |dw:1444091029721:dw|
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