Calculus problem

Question

The position function of a particle in rectilinear motion is given by s(t) = 2t³ - 21t² + 60t + 3 for t ≥ 0 with t measured in seconds and s(t) measured in feet. Find the position and acceleration of the particle at the instant the when the particle reverses direction. Include units in your answer.

Answer 1

The particle reverses direction when the velocity changes from positive to negative. We can find the velocity equation by taking the derivative of the position equation.

s' (t) = 6t²-42t + 60 = v(t)

so, when does v(t) = 0 ? Because it's a quadratic equation, we can solve by factoring, graphing, or using the quadratic formula. I'm going to factor:

6t²-42t + 60 = 0

6(t²-7t + 10) = 0

6(t-2)(t-5) = 0

6=0 t-2 = 0 t-5=0

t = 2, t = 5

The particle changes direction at t = 2 and t = 5. To find the particle's position at these times, evaluate s(2) and s(5). The answer will be in feet.

To find the particle's acceleration at these times, you will have to evaluate a(2) and a(5). However, we don't have an acceleration equation yet. Recall that v'(t) = a(t). Since we know v(t), we can find a(t):

s' (t) = 6t²-42t + 60 = v(t)

v (t) = 6t²-42t + 60

v' (t) = 12t-42 = a(t)

When you evaluate a(2) and a(5), your answer will be in feet/second².