Question

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

\(v\left( t \right) = 1.5\sqrt t ,\;s\left( 4 \right) = 10\)

Short answer

The required position function is \(s\left( t \right) = t\sqrt t + 2\).

Step-by-step solution

Find the anti-derivative of \(v\)

Here we have,\(v\left( t \right) = 1.5\sqrt t \).

To find the position of the particle, we have to find general anti-derivative of velocity.

General anti-derivative of\(v\left( t \right) = 1.5\sqrt t \)I s

\(\begin{aligned}{c}s\left( t \right) &= 1.5\frac{{{t^{{3 \mathord{\left/

{\vphantom {3 2}} \right.

\kern-\nulldelimiterspace} 2}}}}}{{{\raise0.7ex\hbox{$3$} \!\mathord{\left/

{\vphantom {3 2}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$2$}}}} + C\\ &= 1.5\frac{{{t^{{3 \mathord{\left/

{\vphantom {3 2}} \right.

\kern-\nulldelimiterspace} 2}}}}}{{1.5}} + C\end{aligned}\)

Hence, \(s\left( t \right) = t\sqrt t + C\).

Find the value of constant\(C\)

Here we have,\(s\left( t \right) = t\sqrt t + C\).

Also, we have\(s\left( 4 \right) = 10\).

To find value of constant\(C\)put\(s\left( 4 \right) = 10\)in\(s\left( t \right) = t\sqrt t + C\).

\(\begin{aligned}{l} \Rightarrow 10 &= 4\sqrt 4 + C\\ \Rightarrow 10 &= .8 + C\\ \Rightarrow C &= 2\end{aligned}\)

Hence required position function is \(s\left( t \right) = t\sqrt t + 2\).