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andrew11 [14]
3 years ago
6

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?
Physics
1 answer:
Nastasia [14]3 years ago
6 0

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}

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