Question

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

\(v\left( t \right) = \frac{{2t - 1}}{{1 + {t^2}}}\) , \(s\left( 0 \right) = 1\)

Short answer

The position function is, \(s\left( t \right) = \log \left( {1 + {t^2}} \right) - \arctan t + 1\).

Step-by-step solution

Find position function

Given that velocity function is \(v\left( t \right) = \frac{{2t - 1}}{{1 + {t^2}}}\).

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

\(v\left( t \right) = \frac{{2t}}{{1 + {t^2}}} - \frac{1}{{1 + {t^2}}}\)

The general anti-derivative is

\(s\left( t \right) = \log \left( {1 + {t^2}} \right) - \arctan t + C\)

Find the constant

Now, \(s\left( t \right) = \log \left( {1 + {t^2}} \right) - \arctan t + C\)

\(\begin{aligned}{c}s\left( 0 \right) &= \log \left( {1 + {0^2}} \right) - \arctan 0 + C &= 1\\\log 1 - 0 + C &= 1\\0 + C &= 1\\C = 1\end{aligned}\)

Hence the position of particle is \(s\left( t \right) = \log \left( {1 + {t^2}} \right) - \arctan t + 1\).