Question

A particle moves with velocity function v(t) = -t^2 + 5t - 3, with v measured in feet per second and t measured in seconds. Find the acceleration of the particle at time t = 3 seconds.

a. 5/2 feet per second^2
b. -1 feet per second^2
c. -3 feet per second^2
d. 6 feet per second^2

Answer 1

The acceleration of the particle at time t = 3 seconds is found by differentiating the velocity function v(t) = -t^2 + 5t - 3, resulting in an acceleration of -1 feet per second^2.

Step-by-step explanation:

The acceleration of a particle at a given time is found by taking the derivative of the velocity function with respect to time. For the velocity function v(t) = -t^2 + 5t - 3, the acceleration function a(t) is the derivative of v(t), which is a(t) = -2t + 5. To find the acceleration at t = 3 seconds, we substitute 3 into the acceleration function and get a(3) = -2(3) + 5, which simplifies to -1 feet per second^2.