Question

8. A particle's trajectory is described by $x=\left(\frac{1}{2}{t}^3-2t^2\right)\text{m}$and $y=\left(\frac{1}{2}{t}^2-2t\right)\text{m}$, where t is in s.

a. What are the particle's position and speed at $t=0\text{s}$and $t=4\text{s}$?

b. What is the particle's direction of motion, measured as an angle from the x-axis, at $t=0\text{s}$and $t=4\text{s}$?

Short answer

Part (a) Position of particle is zero at t =0 , and t =4 s.

Speed of particle at $t=0\text{s},$$v=2\text{m}/\text{s}$

Speed of particle at t = 4s is$v=8.3\text{m}/\text{s}$

Part (b) the angle is ${14}^{°}\text{northof}+x$

Step-by-step solution

Part (a)

For the position of the aprticle

At $t=0\text{s},x=0\text{m}$and $y=0\text{m}$, or $\overrightarrow{r}=(0\widehat{i}+0\widehat{j})\text{m}$.

At $t=4\text{s},x=0\text{m}$and $\overrightarrow{r}=(0\widehat{i}+0\widehat{j})\text{m}$, or $\overrightarrow{r}=(0\widehat{i}+0\widehat{j})\text{m}$.

In other words, the particle is at the origin at $t=0\text{s}$both and at $t=4\text{s}$

For the velocity of the particle

From the expressions for x and y,

The velocity of the particle is

$\begin{array}{l}\overrightarrow{v}=\frac{dx}{dt}\widehat{i}+\frac{dy}{dt}\widehat{j}\\ =\left[\left(\frac{3}{2}{t}^2-4t\right)\widehat{i}+(t-2)\widehat{j}\right]\text{m}/\text{s}\end{array}$

At $t=0\text{s},\overrightarrow{v}=-2\widehat{j}\text{m}/\text{s},v=2\text{m}/\text{s}$.

At $t=4\text{s},\overrightarrow{v}=(8\widehat{i}+2\widehat{j})\text{m}/\text{s},v=8.3\text{m}/\text{s}$.

Part (b)

(b) At $t=0\text{s},\overrightarrow{v}$is along $t=0\text{s},\overrightarrow{v}$,

or south of +x.

At$t=4\text{s}$ ,

$\theta ={\tan }^{-1}\left(\frac{2\text{m}/\text{s}}{8\text{m}/\text{s}}\right)={14}^{°}\text{northof}+x$