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\,(\mathrm m)/(\mathrm s)`,
`a=-0.37\,(\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\,(\mathrm m)/(\mathrm s)\right)^2=2\left(-0.37\,(\mathrm m)/(\mathrm s^2)\right)(0.79\,\mathrm m)`

`\implies {v_f}^2=0.42\,(\mathrm m^2)/(\mathrm s^2)`

`\implies v_f=0.64\,(\mathrm m)/(\mathrm s)`