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An elevator is moving upward at 1.20 m/s when it experiences an acceleration of 0.31 downward over a distance of 0.75m. What will its final velocity be?

Respuesta :

moonlightbae833
Info.: Vi = 1.20 m/s, a = 0.31 m/s^2, Δx = 0.75 m, Vf = ?

Vf = sqrt (Vi^2 + 2aΔx) = sqrt ((1.20 m/s)^2 + 2(0.31 m/s^2 * 0.75 m))
= 0.99 m/s

Answer:

The final velocity is 0.99 m/s.

Explanation:

Given that,

Speed = 1.20 m/s

Acceleration = -0.31 m/s²

Distance = 0.75 m

We need to calculate the final velocity

Using equation of motion

[tex]v^2=u^2+2as[/tex]

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value in the equation

[tex]v^2=1.20^2-2\times0.31\times0.75[/tex]

[tex]v=\sqrt{0.975}[/tex]

[tex]v=0.99\ m/s[/tex]

Hence, The final velocity is 0.99 m/s.

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