## Related questions with answers

A squirrel has $x$- and $y$-coordinates $(1.1 \mathrm{~m}, 3.4 \mathrm{~m})$ at time
$t_1=0$ and coordinates $(5.3 \mathrm{~m},-0.5 \mathrm{~m})$ at time $t_2=3.0 \mathrm{~s}$. For
this time interval, find

(b) the magnitude and direction of the average velocity.

Solutions

Verified## (b)

We know that the magnitude of the average velocity is given by applying the Pythagorean theorem to the average velocity components. Hence,

$\| v_{avg}\|=\sqrt{v_{x,avg}^2+v_{y,avg}^2}$

Plug from the final red results in part (a) above.

$\| v_{avg}\|=\sqrt{1.4^2+(-1.3)^2}$

$\boxed{\| v_{avg}\|={\color{#c34632}\bf 1.91}\;\rm m/s}$

(b) Average velocity magnitude:

$\left|v_{av}\right|=\sqrt{(v_{av})_x^2+(v_{av})_y^2}=\sqrt{1.4^2+(-1.3)^2}=1.91\,\textrm{ m/s}$

The angle between $x$ and $y$ components of the average velocity:

$\tan{\alpha}=\dfrac{(v_{av})_y}{(v_{av})_x}$

$\alpha=\tan^{-1}{\dfrac{(v_{av})_y}{(v_{av})_x}}=\tan^{-1}{\dfrac{-1.3}{1.4}}=-42.88^{\circ}=317.12^{\circ}$

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