Question

A particle moves on a straight line with velocity function \(v\left( t \right) = sin\omega t\,co{s^2}\omega t\). Find its position function \(s = f\left( t \right)\)if \(f\left( 0 \right) = 0\).

Short answer

The position function of a particle, that moves on a straight line with velocity function \(v\left( t \right) = sin\omega t\,co{s^2}\omega t\) is \( - \frac{{co{s^3}\omega t}}{{3\omega }} + \frac{1}{{3\omega }}\)when \(f\left( 0 \right) = 0\).

Step-by-step solution

Given that a particle moves on a straight line with velocity function\(v\left( t \right) = sin\omega t\,co{s^2}\omega t\). Then, we have its position function as, \(S = \int {v\left( t \right).dt} \)with respect to time.

Consider \(s = f\left( t \right)\)

On substituting,

\(\begin{aligned}{l}s &= f\left( t \right) &= \int {v\left( t \right)} .dt\\ &\Rightarrow \int {sin\omega t.co{s^2}} \omega t.dt\end{aligned}\)

From the trigonometric concept we have,

\(co{s^2}\omega t = 1 - si{n^2}\omega t\)

On substituting and simplifying,

\(\begin{aligned}{l}f\left( t \right) &= \int {sin\omega t\left( {1 - si{n^2}\omega t} \right).dt} \\ &= \int {sin\omega t} \,\,dt - \int {si{n^3}\omega t.dt} \end{aligned}\)

We have,

\(\begin{aligned}{l}{I_\eta } &= \int {si{n^\eta }x.dx} \\ &= \frac{{ - si{n^{\eta - 1}}x.cosx}}{\eta } + \frac{{\eta - 1}}{\eta }.{I_\eta } - 2\end{aligned}\)

On applying above formula, we get,

\(f\left( t \right) = - \frac{{cos\omega t}}{\omega } - \left( { - \frac{{si{n^2}\omega t.cos\omega t}}{{3\omega }} + \frac{2}{3}{I_1}} \right)\)

Again applying the same formula for finding \({I_1}\), we get \({I_1} = \int {sin\omega t\,dt} \). Thus,

\(\begin{aligned}{l}f\left( t \right) &= \frac{{ - cos\omega t}}{\omega } - \left( { - \frac{{si{n^2}\omega t.cos\omega t}}{{3\omega }} + \frac{2}{3}\left( { - \frac{{cos\omega t}}{\omega }} \right)} \right)\\ &= \frac{{ - cos\omega t}}{\omega } + \frac{{si{n^2}\omega t.cos\omega t}}{{3\omega }} + \frac{{2cos\omega t}}{{3\omega }} + c\\ &= \frac{{ - cos\omega t}}{\omega } + \frac{{\left( {1 - co{s^2}\omega t} \right).cos\omega t}}{{3\omega }} + \frac{{2cos\omega t}}{{3\omega }} + c\\ &= \frac{{ - cos\omega t}}{\omega } + \frac{{cos\omega t}}{{3\omega }} - \frac{{co{s^3}\omega t}}{{3\omega }} + \frac{{2cos\omega t}}{{3\omega }} + c\end{aligned}\)

Adding all the common terms,

\(\begin{aligned}{l}f\left( t \right) &= \frac{{ - co{s^3}\omega t}}{{3\omega }} + \frac{{ - 3cos\omega t + cos\omega t + 2cos\omega t}}{{3\omega }} + c\\f\left( t \right) &= \frac{{ - co{s^3}\omega t}}{{3\omega }} + c\end{aligned}\)

Calculating the position function at \(f\left( 0 \right) = 0\).

\(f\left( t \right) = \frac{{ - co{s^3}\omega t}}{{3\omega }} + c\)

\(f\left( 0 \right) = \frac{{ - co{s^3}\omega \left( 0 \right)}}{{3\omega }} + c = 0\)

Therefore, The position function \(s = f\left( t \right)\)if \(f\left( 0 \right) = 0\) is obtained as \(s = f\left( t \right) = \frac{{ - co{s^3}\omega t}}{{3\omega }} + \frac{1}{{3\omega }}\).