Question

Given the position function, s of t equals t cubed divided by 3 minus 15 times t squared divided by 2 plus 50 times t , between t = 0 and t = 12, where s is given in feet and t is measured in seconds, find the interval in seconds where the particle is moving to the left.

Answer 1

s(t) = t^3/3 - 15t^2 / 2 + 50t between t = 0 and t = 12

s' (t) = velocity = t^2 - 15t + 50

The particle is moving to the left when the velocity is negative.

factoring the velocity function we have

v = (t - 10)(t - 5) so the velocity is 0 at t = 5 and t = 10 The graph of this is a parabola with a minimum between t = 5 and 10 ( minimum because of the postive coefficient of t^2) so the require interval when the velocity is negative is between t = 5 and 10.

Answer is its moving to the left for 5 seconds between t = 5 and t = 10 secs.