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.
