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Marta_Voda
4 months ago
10

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 wil

l its final velocity be?
Physics
2 answers:
Maru [3.3K]4 months ago
5 0

Result:

The final velocity computes to 0.99 m/s.

Details:

Given:

Initial speed = 1.20 m/s

Acceleration = -0.31 m/s²

Distance traveled = 0.75 m

Final velocity calculation:

Using the motion equation

v^2=u^2+2as

Where: v = final speed

u = initial speed

a = acceleration

s = distance

Substituting values into the equation yields

v^2=1.20^2-2\times0.31\times0.75

v=\sqrt{0.975}

v=0.99\ m/s

Thus, the final velocity is 0.99 m/s.

kicyunya [3.2K]4 months ago
5 0
<span>Details: Initial velocity, Vi = 1.20 m/s, acceleration, a = 0.31 m/s², distance, Δx = 0.75 m, final velocity, Vf =?

To find Vf, we use the formula: Vf = sqrt (Vi^2 + 2aΔx) = sqrt ((1.20 m/s)^2 + 2(0.31 m/s² * 0.75 m))
= 0.99 m/s</span>
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