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

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You need to integrate the given velocity function to find the position of particle such that:

`s(t) = int (sin alpha*t*cos alpha*t)^2 dt`

You need to use the half angle identities such that:

`sin^2 alpha*t = (1 - cos 2alpha*t)/2`

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

`s(t) = int (1 - cos 2alpha*t)/2*(1 + cos 2alpha*t)/2 dt`

`s(t) = (1/4)int (1 - cos 2alpha*t)(1 +cos 2alpha*t) dt`

You need to convert the integrand into a difference of squares such that:

`s(t) = (1/4)int (1 - cos^2 (2alpha*t)) dt`

Using the property of linearity such that:

`s(t) = (1/4)int dt- (1/4)int (cos^2 (2alpha*t)) dt`

`cos^2 (2alpha*t) = (1 + cos 4alpha*t)/2`

`s(t) = (1/4)int dt- (1/4)int(1 + cos 4alpha*t)/2 dt`

`s(t) = (1/4)int dt - (1/8)int (1 + cos 4alpha*t) dt`

`s(t) = (1/4)int dt - (1/8)int dt - (1/8)int (cos (4alpha*t)) dt`

`s(t) = (1/8)int dt - (1/8)int (cos (4alpha*t)) dt`

`s(t) = (1/8)(t - (sin (4alpha*t))/4) + c`

The problem provides the information that `s(0) = 2` , hence, you may evaluate the constant `c` such that:

`s(0) = (1/8)(0 - (sin (0))/4) + c`

Since si`n 0 = 0 ` yields:

`s(0) = c => c = 2`

**Hence, evaluating the position of the particle, under the given conditions, yields `s(t) = (1/8)(t - (sin (4alpha*t))/4) + 2` .**

V(t)

`= sinalphat*(cosalphat)^2`

Position function is given by the integration of vector function.

S(t)

`= int V(t)dt`

`= intsinalphat*(cosalphat)^2*dt`

Let

`x = cosalphat`

`(dx)/dt = -sinalphat*alpha`

`dx = -alpha*sinalphat*dt`

S(t)

`= intsinalphat*(cosalphat)^2*dt`

`= int -alpha*x^2dx`

`= -alpha int x^2dx`

`= -alpha(x^3/3)+C`

`= -alpha((cosalphat)^3/3)+C`

It is given that S(0) = 2

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

`2 = -alpha/3+C`

`C = 2+alpha/3`

`S(t) = -alpha((cosalphat)^3/3)+2+alpha/3`

*So the position function is;*

`S(t) = (alpha/3)(1-(cosalphat)^3)+2`

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