Question

The figure gives the path of a squirrel moving about on level ground, from point A (at time t-O), to points B (at t = 5 min), C (at t = 10 min), and finally D (att-15 min). Consider the average velocities of the squirrel from point A to each of the other three points. Of them, what are the (a) magnitude and (b) angle of the one with the least magnitude and (c) the magnitude and (d) angle of the one with the greatest magnitude? Give the angles in the range (-1809, 180°] y (m) 50 25 25 50 -25 -50

Answer 1

Average velocity $\vec{v}_{avg}=\frac{\text{displacement}}{\text{time}}=\frac{\Delta\vec{r}}{\Delta t}$

$\vec{r}_A=(15\hat{x}-15\hat{y})\,m$  $t_A=0\,min$

$\vec{r}_B=(30\hat{x}-45\hat{y})\,m$  $t_B=5\,min$

$\vec{r}_C=(20\hat{x}-15\hat{y})\,m$  $t_C=10\,min$

$\vec{r}_D=(45\hat{x}+45\hat{y})\,m$  $t_D=15\,min$

Average velocities are

$\vec{v}_{BA}=\frac{\vec{r}_B-\vec{r}_A}{t_B-t_A}=\frac{(30\hat{x}-45\hat{y})-(15\hat{x}-15\hat{y})}{5-0}=(3\hat{x}-6\hat{y})\,m/min$

Magnitude $|\vec{v}_{BA}|=\sqrt{3^2+6^2}=3\sqrt{5}\,m/min$

Direction $\theta_{BA}=-\tan^{-1}{\left(\frac{6}{3} \right )}=-63.4\degree$

$\vec{v}_{CA}=\frac{\vec{r}_C-\vec{r}_A}{t_C-t_A}=\frac{(20\hat{x}-15\hat{y})-(15\hat{x}-15\hat{y})}{10-0}=0.5\hat{x}\,m/min$

Magnitude $|\vec{v}_{CA}|=0.5\,m/min$

Direction $\theta_{CA}=0\degree$

$\vec{v}_{DA}=\frac{(45\hat{x}+45\hat{y})-(15\hat{x}-15\hat{y}))}{15-0}=(2\hat{x}+4\hat{y})\,m/min$

Magnitude $|\vec{v}_{DA}|=\sqrt{2^2+4^2}=2\sqrt{5}\,m/min$

Direction $\theta_{DA}=\tan^{-1}{\left(\frac{4}{2} \right )}=63.4\degree$

(a)

Smallest magnitude $|\vec{v}_{CA}|=0.5\,m/min$

(b)

$\theta_{CA}=0\degree$

(c)

$|\vec{v}_{BA}|=3\sqrt{5}\,m/min$

(d)

$\theta_{BA}=-63.4\degree$

Answer 2

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