Hello sderosa518 Originally Posted by

**sderosa518** A particle moves that its position vector r at the time t is given by

r = c ( t - 1/3 t^3) i + ct^2 j

Where c is a positive constant.

Find the velocity at time t(> 0) and show that’s its magnitude increase with time.

Determine its direction of motion at time t=0 show that the directions turns through a right angle between t =0 and t=1.

Show also that the component of force acting on the particle in the direction of the vector j is constant throughout the motion.

Given that c=1, find the maximum displacement of the particle in the I direction and draw a rough sketch of the path of the particle for values of t> 0, showing clearly the positions of the particle at times t=0, 1, radical 3, 2.

$\displaystyle \textbf{r} = c(t-\tfrac13t^3)\textbf{i}+ct^2\textbf{j}$

$\displaystyle \Rightarrow \textbf{v}=\frac{d\textbf{r}}{dt}=c(1-t^2)\textbf{i}+2ct\textbf{j}$

The magnitude of the velocity $\displaystyle = \sqrt{c^2(1-t^2)^2+4c^2t^2}$$\displaystyle =c\sqrt{1-2t^2+t^4+4t^2}$

$\displaystyle =c\sqrt{1+2t^2+t^4}$

$\displaystyle =c\sqrt{(1+t^2)^2}$

$\displaystyle =c(1+t^2)$

which clearly increases as $\displaystyle t$ increases, for $\displaystyle t > 0$, since $\displaystyle c$ is a positive constant.

When $\displaystyle t = 0,\; \textbf{v}= c\textbf{i}$. So the direction of motion when $\displaystyle t=0$ is parallel to the vector $\displaystyle \textbf{i}$ (i.e. parallel to the $\displaystyle x$-axis).

When $\displaystyle t = 1,\; \textbf{v}= 2c\textbf{j}$. So the direction of motion when $\displaystyle t=1$ is parallel to the vector $\displaystyle \textbf{j}$ (i.e. parallel to the $\displaystyle y$-axis), and is therefore perpendicular to its direction when $\displaystyle t = 0$.

Acceleration $\displaystyle = \textbf{a} = \frac{d\textbf{v}}{dt} = -2ct\textbf{i}+2c\textbf{j}$. The component in the $\displaystyle \textbf{j}$ direction is therefore constant. So the component of the external force that causes this acceleration is also constant in this direction.

The maximum displacement in the $\displaystyle \textbf{i}$ direction will occur when the component of the velocity in this direction is zero; i.e. when $\displaystyle c(1-t^2)=0$; i.e. when $\displaystyle t = 1$ (for $\displaystyle t>0$)

So when $\displaystyle t = 1$ and $\displaystyle c = 1$, the maximum displacement in the $\displaystyle \textbf{i}$ direction is $\displaystyle 1(1-\tfrac13.1^3) = \tfrac23$.

I'll leave you to work out the sketch of the path of the particle. Simply give $\displaystyle t$ various sensible values (including those given in the question) and work out the position vector $\displaystyle \textbf{r}$ for each value. Plot these points to sketch the path.

Grandad