Chapter 3, Problem 1FP

To determine

The position of the particle.

Explanation of Solution

Determine the position of the particle (x, y).

  $\frac{dx}{dt}=u=\frac{1}{4}x$

Here, velocity component in x location is u.

Integrate on both sides.

$\begin{array}{l}{\displaystyle \underset{2}{\overset{x}{\int }}\frac{dx}{x}}={\displaystyle \underset{0}{\overset{t}{\int }}\frac{1}{4}}dt\\ {\left(\ln x\right)}_2^x={\left(\frac{t}{4}\right)}_0^t\\ \ln x-\ln 2=\frac{t}{4}-0\end{array}$

$\begin{array}{l}\ln \frac{x}{2}=\frac{t}{4}\\ \frac{x}{2}=e^{\frac{t}{4}}\end{array}$

$\begin{array}{c}x=2e^{\frac{t}{4}}\\ x=2e^{\frac{2\text{s}}{4}}\\ x=3.30\text{m}\end{array}$

Determine the position of the particle (x, y).

  $\frac{dy}{dt}=v=2t$

Here, velocity component in y location is v.

Integrate on both sides.

$\begin{array}{l}{\displaystyle \underset{6}{\overset{y}{\int }}dy}={\displaystyle \underset{0}{\overset{t}{\int }}2t}dt\\ {\left(y\right)}_6^y={\left(2\times \frac{t^2}{2}\right)}_0^t\\ y-6=t^2-0\end{array}$

$\begin{array}{c}y=t^2+6\\ y={\left(2\text{s}\right)}^2+6\\ y=10\text{m}\end{array}$

Thus, the position of the particle (x, y) is $\left(3.3,10\right)$.