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) = (\Delta \vec r)/(\Delta t) = ((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)(\rm m)/(\rm s)`

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

`-2.25(\rm m)/(\rm s) = (x_2 - (-17.5\,\mathrm m))/(10.7\,\mathrm s) \\\\ 1.79(\rm m)/(\rm s) = (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(\rm m)/(\rm s)\right)(10.7\,\mathrm s) \approx \boxed{-6.58\,\mathrm m}`

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