Question

Question #cd675

Answer 1

`v(t) \equiv \frac{ds}{dt}; \qquad \int_{s(0)}^{s(t)}ds = \int_0^t v(t) dt`
`s(t) - s(0) = \int_0^t (40 - \sint) dt = 40t + [cos(t)]_0^t`
`s(t) = s(0) + 40t + (\cost - 1) = 1 + \cost + 40t`

Answer 2

The answer is
`s(t)=40t+cos(t)+1`

Explanation:

The position function is the intehral of the velocity function

`v(t)=40-sin(t)`

`s(t)=int(40-sin(t))dt=40t+cos(t)+C`

Plugging in the initial conditions, `s(0)=2`

Therefore,

`s(0)=40*0+cos(0)+C`

`2=1+C`

`C=1`

`s(t)=40t+cos(t)+1`