Question

A puck moves 2.35 m/s in a -22° direction. A hockey stick pushes it for 0.215 s, changing its velocity to 6.42 m/s in a 50.0° direction. What was the direction of the acceleration?

Answer 1

The puck starts with velocity vector

$\vec v_0=\left(2.35\dfrac{\rm m}{\rm s}\right)(\cos(-22^\circ)\,\vec\imath+\sin(-22^\circ)\,\vec\jmath)=(2.18\,\vec\imath-0.880\,\vec\jmath)\dfrac{\rm m}{\rm s}$

Its velocity at time $t$ is

$\vec v=\vec v_0+\vec at$

Over the 0.215 s interval, the velocity changes to

$\vec v=\left(6.42\dfrac{\rm m}{\rm s}\right)(\cos50.0^\circ\,\vec\imath+\sin50.0^\circ\,\vec\jmath)=(4.13\,\vec\imath+4.92\,\vec\jmath)\dfrac{\rm m}{\rm s}$

Then the acceleration must have been

$\vec v=\vec v_0+(0.215\,\mathrm s)\vec a\implies\vec a=\dfrac{\vec v-\vec v_0}{0.215\,\rm s}=(9.06\,\vec\imath+27.0\,\vec\jmath)\dfrac{\rm m}{\mathrm s^2}$

which has a direction of about $71.4^\circ$.