Question 1178329: A particle P travels in a straight line and its distance, s metres, from a fixed point O, is given by s= t^3 - 9t^2 + 15t + 40, where t is the time in seconds after passing O. Calculate:
a) the distance of P from O when it is instantaneously at rest
b) the values of t when the acceleration has a magnitude of 12 ms
c) the average velocity of P during the first 2 seconds
d) the total distance travelled in the first 6 seconds
Answer by Solver92311(821)   (Show Source):
You can put this solution on YOUR website!
\ =\ t^3\ -\ 9t^2\ +\ 15t\ +\ 40)
\ =\ s'(t)\ =\ \frac{ds}{dt})
\ =\ 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
My calculator said it, I believe it, that settles it
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