Question

A particle is moving with the given data. Find the position of the particle

\(a\left( t \right) = sint + 3cost\) , \(s\left( 0 \right) = 0,v\left( 0 \right) = 2\)

Short answer

The position function is, \(s\left( t \right) = \sin t - 3\cos t + 3t + 3\).

Step-by-step solution

Find velocity function

Given that acceleration function is \(a\left( t \right) = \sin t + 3\cos t\)

The velocity function is anti-derivative of acceleration function.

The anti-derivative is

\(v\left( t \right) = - \cos t + 3\sin t + C\)

Since \(v\left( 0 \right) = 2\)

\(\begin{aligned}{c} - \cos 0 + 3\sin 0 + C &= 2\\ - 1 + 0 + C &= 2\\C &= 3\end{aligned}\)

 Step 2: Find position function

Now velocity function is \(v\left( t \right) = - \cos t + 3\sin t + 3\).

The position function is anti-derivative of the velocity function.

The general anti-derivative is

\(s\left( t \right) = \sin t - 3\cos t + 3t + C\)

Now, \(s\left( 0 \right) = 0\)

\(\begin{aligned}{c}s\left( 0 \right) &= \sin 0 - 3\cos 0 + 3\left( 0 \right) + C &= 0\\0 - 3 + 0 + C &= 0\\C &= 3\end{aligned}\)

Hence the position of particle is \(s\left( t \right) = \sin t - 3\cos t + 3t + 3\).