A particle moves with velocity function v(t) = 2t^2 − 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.

a)3/4 feet per second^2
b)-1 feet per second^2
c)32 feet per second^2
d)5 feet per second^2

a = 4 t -3

s(2) = 8-3 = 5

To find the acceleration of the particle at time t = 2 seconds, we need to find the derivative of the velocity function v(t).

The derivative of a function represents the rate of change of that function at any given point. In this case, we want to find the rate of change of velocity with respect to time, which gives us acceleration.

The derivative of v(t) is denoted as v'(t) or dv/dt, and it can be found by differentiating each term of the velocity function with respect to t.

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

Differentiate each term:
v'(t) = d/dt (2t^2) - d/dt (3t) - d/dt (3)

The derivative of t^n (where n is any constant) is n*t^(n-1), so the derivatives of the terms are as follows:

v'(t) = 2 * (d/dt (t^2)) - 3 * (d/dt (t)) - 0

v'(t) = 2 * 2t - 3 * 1

v'(t) = 4t - 3

Now we have the expression for the acceleration, v'(t) = 4t - 3.

To find the acceleration at t = 2 seconds, substitute t = 2 into the expression:

a(2) = 4(2) - 3
a(2) = 8 - 3
a(2) = 5

Therefore, the acceleration of the particle at time t = 2 seconds is 5 feet per second^2. So, the correct answer is option d) 5 feet per second^2.