# A particle moves on a straight line with velocity function `v(t)= sin(alpha t)cos^2(alpha t)` . Find its position function s(t) if s(0) = 2.

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### 1 Answer

Note that,

`s(t) = int v dt`

So, substitute `v(t) = sin(alpha t)cos^2(alpha t) ` to the formula above.

`s(t) = intsin(alpha t)cos^2(alpha t) dt`

To integrate, apply u-substitution.

Let,

`u = cos (alpha t)`

Then, differentiate u.

`du = -sin (alpha t) * alpha dt`

`-(du)/alpha = sin (alpha t) dt`

Replacing `cos (alpha t)` with `u` and `sin (alpha t)dt` with `-(du)/alpha` yields,

`s(u) = int u^2 * (-du)/alpha = -1/alpha int u^2 du`

Then, apply the power formula of integral which is `int u^n du = u^(n+1)/(n+1) + C ` .

`s(u) = -1/alpha u^3/3 + C`

`s(u)= -u^3/(3alpha) + C`

Substitute back `u= cos (alpha t)` to return to the function s(t).

`s(t) = -(cos^3 (alpha t))/(3alpha) + C`

To determine the value of C, use the given condition s(0) = 2

`2 = -(cos^3(alpha*0))/(3alpha)+ C`

`2=-1/(3alpha)+C`

`1/(3alpha)+ 2=C`

**Hence the position function is:**

**`s(t) = -(cos^3(alpha t))/(3 alpha) + 1/(3alpha)+2` **