That is correct.
a(u) = 32
v(u) = 32u + Vo
v(t) - v(0) = 32t
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!!
a(u) = 32
v(u) = 32u + Vo
v(t) - v(0) = 32t
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).