Question 1178329
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ s(t)\ =\ t^3\ -\ 9t^2\ +\ 15t\ +\ 40]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ v(t)\ =\ s'(t)\ =\ \frac{ds}{dt}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(t)\ =\ v'(t)\ = \frac{dv}{dt}\ =\ \frac{d^2s}{dt^2}]


a. Calculate s', set s' equal to zero, solve for the two values of t, then evaluate s at the two values of t.


b. Cannot do this as written.  Acceleration is measured in meters per second per second.  Acceleration = 12 ms is meaningless.  Presuming that you meant the proper units, calculate v', set v' equal to 12, solve for t.


c. Evaluate v(t) at 0 and at 2, calculate the difference in velocity between the two times, divide by the elapsed time.


d. Evaluate the definite integral of s(t)dt from 0 to 6.


																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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