Question

A particle moving at a velocity of 4.5 m/s in the positive x direction is given an acceleration of 7.1 m/s 2 in the positive y direction for 0.7 s. What is the final speed of the particle? Answer in units of m/s.

Answer 1

The particle's velocity in the
`y` direction at time
`t` is

`v_y(t)=\left(7.1\,(\mathrm m)/(\mathrm s^2)\right)t`

so that after 0.7 s, its velocity in the
`y` direction is

`v_y=4.97\,(\mathrm m)/(\mathrm s)`

At this point, the particle has a velocity vector
`v=\langle v_x,v_y\rangle=\langle4.5,4.97\rangle\,(\mathrm m)/(\mathrm s)`. Its speed will be the magnitude of this vector, which is given by

`\|v\|=\sqrt{\left(4.5\,(\mathrm m)/(\mathrm s)\right)^2+\left(4.97\,(\mathrm m)/(\mathrm s)\right)^2}\approx6.7\,(\mathrm m)/(\mathrm s)`