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 $v_f$:

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