Question

A particle starts to move in a straight line from

Answer 1

Answer: s(5) = -200,  v(5) = -90

              s(10) = -900, v(10) = -190

Explanation:

Position: s(t)

Velocity:  s'(t) = v(t)                ⇒      $s(t)=\int {v(t)} \, dt$

Acceleration     v'(t) = a(t)      ⇒      $v(t)=\int {a(t)} \, dt$

We are given that acceleration a(t) = -20   and velocity v(t) = 10

$v(t)=\int {a(t)} \, dt\\\\v(t)=\int{-20}\, dt\\\\v(t)=-20t + C \\\\v(t)=10\quad \longrightarrow \quad C=10\\\\v(t)=-20t+10$

$s(t)=\int {v(t)} \, dt\\\\s(t)=\int {(-20t+10)} \, dt\\\\s(t)=-10t^2+10t$

(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