Calculus I practice help, moving object problem

Question

Suppose the position of an object moving horizontally after t seconds is given by the following function

s = 2t^3 − 3t^2 + t, where s is measured in meters. What is the average velocity on the interval [0, 2]? What is the velocity and the acceleration of the object when t = 1. For what values of t is the velocity zero?

Answer 1

If the position s is given by s(t) = 2t³ − 3t² + t, then the average velocity can be calculated by the delta distance traveled divided by the delta time or [s(2) - s(0)]/(2) = (6 - 0)/2 = 3 m/s

The instantaneous velocity at t = 1 is given by ds/dt. Taking the derivative of s gives s'(t) = 6t² - 6t + 1 and s'(1) = 1 m/s. The instantaneous acceleration at t = 1 is given by the second derivative d²s/dt². Taking the derivative of s' gives s''(t) = 12t - 6 and s''(1) = 6 m/s²

If we want to know when v(t) = 0 then we set the velocity function equal to zero and solve:

s'(t) = v(t) = 6t² - 6t + 1

6t² - 6t + 1 = 0

Using the quadratic formula, t = [6 ± √(36 - 4(6)(1)]/[(2)(6)] = [6 ± √12]/12 or

t = 0.2113 s and t = 0.7887 s