The net change in velocity during the given time interval is equal to -0.665.
Given the following data:
Since acceleration is the rate of change of velocity of an object, we would derive an expression for the velocity by differentiating with respect to time:
[tex]V(t) = \frac{d}{dt} a(t)\\\\V(t) = \frac{d}{dt} (sin(\frac{t^2}{3} ))\\\\V(t) = cos( \frac{t^2}{3})(\frac{2t}{3})\\\\V(t) = \frac{2t}{3}cos( \frac{t^2}{3})[/tex]
Substituting the time interval, we have:
[tex]V(t) = \frac{2\times 2.25}{3}cos( \frac{2.25^2}{3})- \frac{2\times 0.75}{3}cos( \frac{0.75^2}{3})\\\\V(t) = \frac{4.5}{3}cos( \frac{5.0625}{3})- \frac{1.5}{3}cos( \frac{0.5625}{3})\\\\V(t) =1.5(-0.11643)-0.5(0.98247)[/tex]
V(t) = -0.665.
Read more on acceleration here: brainly.com/question/24728358
