Question

The position vector $\overrightarrow{r}=5.00t\hat{i}+(et+f{t}^2)\hat{j}$ locates a particle as a function of time t. Vector $\overrightarrow{r}$ is in meters, t is in seconds, and factors e and f are constants. Below Diagram gives the angle $\theta$ of the particle's direction of travel as a function of t($\theta$) is measured from the positive x direction). What are 'e' and 'f', including units?

• A. $e=\frac{35}{6\sqrt{3}}m/s,f=\frac{-5}{24\sqrt{3}}m/s^2$
• B. $e=\frac{-35}{6\sqrt{3}}m/s,f=\frac{5}{24\sqrt{3}}m/s$
• C. $e=\frac{35}{6\sqrt{3}}m/s,f=\frac{5}{24\sqrt{3}}m/s^2$
• D. $e=\frac{35}{6\sqrt{3}}m/s,f=\frac{5}{24\sqrt{3}}m/s^2$

Answer 1

The correct option is A

$e=\frac{35}{6\sqrt{3}}m/s,f=\frac{-5}{24\sqrt{3}}m/s^2$

Given the graph between $\theta$ and t you can write the equation of line using straight line equation,
y=mx+c
$\theta =\frac{-5}{2}t+35$
Given that this $\theta$ is the particles direction of travel.
We know direction of travel is the direction of velocity vector and not position vector.
So let's find velocity by differentiating
$r=5t\hat{i}+(et+f{t}^2)\hat{j}$
$v=5\hat{i}+(e+2ft)\hat{j}$
$\therefore v_x=5m/s{v}_y=(e+2ft)m/s$.

Now tan $\theta$ will always be $\frac{v_y}{v_x}$

From the graph, we can see that
At t=14, $\theta =0,\therefore tan\left(\theta \right)=0$

$\Rightarrow att=14s,\frac{v_y}{v_x}=0\Rightarrow v_y=0$
$v_y=e+2ft$
$\Rightarrow e+2f\left(14\right)=0$
$\Rightarrow e+28f=0......\left(1\right)$
at t= 2 s, $\theta =30$
$\Rightarrow -24f=\frac{5}{\sqrt{3}}$
$\Rightarrow f=\frac{-5}{24\sqrt{3}}$
$e=-28f=-28\times \left(\frac{-5}{24\sqrt{3}}\right)$
$=\frac{7}{6}\times \frac{5}{\sqrt{3}}=\frac{35}{6\sqrt{3}}$

To find units of e and f, we know
$v_y=(e+2ft)m/s$
$\therefore$ e must have a unit of m/s and ft must have a unit of m/s.
$\Rightarrow$ f must have a unit of $m/s^2$ as t has a unit of 's'

$\therefore e=\frac{35}{6\sqrt{3}}m/sandf=\frac{-5}{24\sqrt{3}}m/s^2$