# A point object is moving on the cartesian coordinate plane according to: vecr(t)=b^2thatu_x+(ct^3−q_0)hatu_y. Determine: a) The equation of the trajectory of object on the cartesian plane b) The magnitude and direction of the velocity?

Aug 8, 2016

Let $\left(x , y\right)$ represents the cartesian coordinate of the point having position vector at t th instant

$\vec{r} \left(t\right) = {b}^{2} t {\hat{u}}_{x} + \left(c {t}^{3} - {q}_{0}\right) {\hat{u}}_{y} \ldots . \left(1\right)$

Then
$x = {b}^{2} t \mathmr{and} y = c {t}^{3} - {q}_{0}$

$\therefore y = c {\left(\frac{x}{b} ^ 2\right)}^{3} - {q}_{0}$

On simplification we get

${b}^{6} y = c {x}^{3} - {q}_{0} {b}^{6}$
$\implies c {x}^{3} - {b}^{6} y - {q}_{0} {b}^{6} = 0 \to$ is the eqution of trajectory.

Now differentiating (1) w.r.to.t we get velocity

$\vec{v} \left(t\right) = \frac{d \left(r \left(t\right)\right)}{\mathrm{dt}} = {b}^{2} {\hat{u}}_{x} + 3 c {t}^{2} {\hat{u}}_{y}$
Its magnitude

$\left\mid \vec{v} \left(t\right) \right\mid = \sqrt{{b}^{4} + 9 {c}^{2} {t}^{4}}$

The direction of the velocity.
If it is directed at an angle $\theta \left(t\right)$ with the x-axis at t th instant then

$\theta \left(t\right) = {\tan}^{-} 1 \left(\frac{3 c {t}^{2}}{b} ^ 2\right)$