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 second2
b)-1 feet per second2
c) -3 feet per second2
d) 6 feet per second2

Answer 1

The acceleration of the particle at t = 3 seconds is calculated by finding the derivative of the velocity function, which results in -1 foot per second squared.

Step-by-step explanation:

To find the acceleration of the particle at time t = 3 seconds, we must first understand that acceleration is the derivative of the velocity function with respect to time. The velocity function given is v(t) = -t2 + 5t - 3. To find the acceleration function a(t), we take the derivative of v(t).

a(t) = v'(t) = d(-t2 + 5t - 3)/dt = d(-t2)/dt + d(5t)/dt - d(3)/dt = -2t + 5 + 0 = -2t + 5.

Now, we substitute t = 3 seconds into the acceleration function to find the acceleration at that specific time:

a(3 s) = -2(3) + 5 = -6 + 5 = -1.

The acceleration of the particle at t = 3 seconds is -1 foot per second2.