Recall that the acceleration a(t) of a particle moving along a straight line is the instantaneous rate of change of the velocity v(t); that is,

a(t) = d/dt v(t)

Assume that a(t) = 32 ft/s². Express the cumulative change in velocity during the interval [0, t] as a definite integral, and compute the integral.

So far I got:

v(t) - v(0) = the integral from 0 to t of a(u)du, but im not sure where to go from there.

Thank you so much for your help!!

That is correct.

a(u) = 32

v(u) = 32u + Vo

v(t) - v(0) = 32t

To express the cumulative change in velocity during the interval [0, t] as a definite integral, you are on the right track. The formula you have is correct. Here's how you can proceed:

Step 1: Start with the given formula:
v(t) - v(0) = ∫[0, t] a(u) du

Step 2: Substitute the given value of acceleration:
v(t) - v(0) = ∫[0, t] 32 ft/s² du

Step 3: Integrate with respect to u:
v(t) - v(0) = 32 ∫[0, t] du

Step 4: Evaluate the definite integral:
v(t) - v(0) = 32 [u] from 0 to t

Step 5: Substitute the limits of integration:
v(t) - v(0) = 32 (t - 0)

Step 6: Simplify the expression:
v(t) - v(0) = 32t

Step 7: Solve for v(t) to get the cumulative change in velocity during the interval [0, t]:
v(t) = 32t + v(0)

Therefore, the cumulative change in velocity during the interval [0, t] is given by the equation v(t) = 32t + v(0).