Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises - Page 238: 89

$-16t^{2}+20t$

v(t)= $\int a(t)dt= \int -32dt = -32\int dt$= -32t+C. v(0)= 20 implies C= 20. That is v(t)= -32t+20. Now s(t)=$ \int v(t)dt= \int (-32t+20)dt$ $= -32\int tdt + 20\int dt=-32\frac{t^{2}}{2}+20t+C$ $=-16t^{2}+20t+C$. s(0)= 0 implies that C=0. Thus we get, s(t)= $-16t^{2}+20t$.