Answer: s(5) = -200, v(5) = -90
s(10) = -900, v(10) = -190
Explanation:
Position: s(t)
Velocity: s'(t) = v(t) ⇒ [tex]s(t)=\int {v(t)} \, dt[/tex]
Acceleration v'(t) = a(t) ⇒ [tex]v(t)=\int {a(t)} \, dt[/tex]
We are given that acceleration a(t) = -20 and velocity v(t) = 10
[tex]v(t)=\int {a(t)} \, dt\\\\v(t)=\int{-20}\, dt\\\\v(t)=-20t + C \\\\v(t)=10\quad \longrightarrow \quad C=10\\\\v(t)=-20t+10[/tex]
[tex]s(t)=\int {v(t)} \, dt\\\\s(t)=\int {(-20t+10)} \, dt\\\\s(t)=-10t^2+10t\\\\[/tex]
(a) Input t = 5 into the s(t) and v(t) equations
s(5) = -10(5)² + 10(5) v(5) = -20(5) + 10
= -250 + 50 = -100 + 10
= -200 = -90
(b) Input t = 10 into the s(t) and v(t) equations
s(10) = -10(10)² + 10(10) v(10) = -20(10) + 10
= -1000 + 100 = -200 + 10
= -900 = -190
