Question

An elevator is moving upward at 1.00 m/s when it experiences an acceleration 0.37 m/s2 downward, over a distance of 0.79 m. What will its final velocity be?

Answer 1

Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

We're given $v_i=1.00\,\frac{\mathrm m}{\mathrm s}$, $a=-0.37\,\frac{\mathrm m}{\mathrm s^2}$ (so we take the upward direction to be positive), and $\Delta y=0.79\,\mathrm m$. Then the final velocity $v_f$ satisfies

${v_f}^2-\left(1.00\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-0.37\,\dfrac{\mathrm m}{\mathrm s^2}\right)(0.79\,\mathrm m)$

$\implies {v_f}^2=0.42\,\dfrac{\mathrm m^2}{\mathrm s^2}$

$\implies v_f=0.64\,\dfrac{\mathrm m}{\mathrm s}$