Question

Starting from a location with position vector 1,=−17.5 m and 1,=23.1 m , a rabbit hops around for 10.7 seconds with average velocity ,=−2.25 m/s and ,=1.79 m/s . Find the components of the position vector of the rabbit's final location, 2, and 2, .

Answer 1

By definition of average velocity,

$\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t} = \dfrac{(x_2-x_1)\,\vec\imath + (y_2-y_1)\,\vec\jmath}{10.7\,\mathrm s}$

If

$\vec v_{\rm ave} = (-2.25\,\vec\imath + 1.79\,\vec\jmath)\dfrac{\rm m}{\rm s}$

and $x_1=-17.5\,\mathrm m$ and $y_1=23.1\,\mathrm m$, then

$-2.25\dfrac{\rm m}{\rm s} = \dfrac{x_2 - (-17.5\,\mathrm m)}{10.7\,\mathrm s} \\\\ 1.79\dfrac{\rm m}{\rm s} = \dfrac{y_2 - 23.1\,\mathrm m}{10.7\,\mathrm s}$

Solve for $x_2$ and $y_2$:

$x_2 = 17.5\,\mathrm m + \left(-2.25\dfrac{\rm m}{\rm s}\right)(10.7\,\mathrm s) \approx \boxed{-6.58\,\mathrm m}$

$y_2 = 23.1\,\mathrm m + \left(1.79\dfrac{\rm m}{\rm s}\right)(10.7\,\mathrm s) \approx \boxed{42.3\,\mathrm m}$