Question

A particle moves with velocity function v(t) = 2t2 - 3t - 3, with v measured in feet per second and t measured in seconds. Find the acceleration of the particle at time t = 2 seconds. (4 points)

a. 3/4 ft/s^2
b. -1 ft/s^2
c. 32ft/s^2
d. 5 ft/s^2

Answer 1

d. 5 ft/s^2

Explanation:

The acceleration is the derivative of the velocity.

In this problem, we have the velocity function

`v(t) = 2t^(2) - 3t -3`

So the acceleration function is:

`a(t) = v^(\prime)(t) = 4t - 3`.

Find the acceleration of the particle at time t = 2 seconds

This is a(2). So:

`a(2) = v^(\prime)(2) = 4(2) - 3 = 5`.

The correct answer is

d. 5 ft/s^2