Question

I need help with this physics question, "A skateboarder rolled down the sidewalk with an initial velocity of 2.5m/s. If her acceleration was 1.7 m/s^2, what was her velocity after she had skated 130m?"

Answer 1

Recall that

`{v_f}^2={v_i}^2+2a\Delta x`

where
`v_i` and
`v_f` are the skateboarder's velocity, respectively;
`a` is her acceleration; and
`\Delta x` is the change in her position.

Substitute everything you know and solve for
`v_f`:

`{v_f}^2=\left(2.5(\rm m)/(\rm s)\right)^2+2\left(1.7(\rm m)/(\mathrm s^2)\right)(130\,\mathrm m)\implies v_f=448.25(\mathrm m^2)/(\mathrm s^2)\implies\boxed{v_f\approx21(\rm m)/(\rm s)}`