# 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?

Jun 4, 2016

trajectory $y = \left(\frac{c}{{b}^{6}}\right) {x}^{3} - {q}_{0}$
velocity magnitude $\left\lVert \vec{v} \left(t\right) \right\rVert = \sqrt{{b}^{4} + 9 {c}^{2} {t}^{4}}$
velocity normalized direction $\frac{\vec{v} \left(t\right)}{\left\lVert \vec{v} \left(t\right) \right\rVert} = \frac{\left\{{b}^{2} , 3 c {t}^{2}\right\}}{\sqrt{{b}^{4} + 9 {c}^{2} {t}^{4}}}$

#### Explanation:

Making $\vec{r} \left(t\right) = \left\{x \left(t\right) , y \left(t\right)\right\}$ we have

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

which is the so called movement parametric equation.
Eliminating $t$ we get

$y = \left(\frac{c}{{b}^{6}}\right) {x}^{3} - {q}_{0}$

which is the trajectory equation.

Velocity determination is done differentiating $\vec{r} \left(t\right)$ regarding $t$ so

$\vec{v} \left(t\right) = \left\{{b}^{2} , 3 c {t}^{2}\right\}$

so $\left\lVert \vec{v} \left(t\right) \right\rVert = \sqrt{{b}^{4} + 9 {c}^{2} {t}^{4}}$ is the velocity magnitude and

$\frac{\vec{v} \left(t\right)}{\left\lVert \vec{v} \left(t\right) \right\rVert} = \frac{\left\{{b}^{2} , 3 c {t}^{2}\right\}}{\sqrt{{b}^{4} + 9 {c}^{2} {t}^{4}}}$ is the normalized direction