An object moving with uniform acceleration has a velocity of 10.0 cm/s in the positive x-direction when its x-coordinate is 3.18 cm. if its x-coordinate 2.85 s later is −5.00 cm, what is its acceleration?

For this case we have the following equation:

Δx = ½ a (Δt) ² + vΔt

Where,

Δx: displacement

a: acceleration

Δt: total time

v: initial speed

Clearing the acceleration we know:

a = 2 (Δx - vΔt) / (Δt) ²

We now perform the calculations.

The total time is:

Δt = 2.85 - 0 = 2.85

Δx = - 5 - 3.18 = - 8.18

v = 10

Substituting values we have:

a = 2 (-8.18 - (10) (2.85)) / (2.85) ²

a = -9.03 cm / s ^ 2

The acceleration of the object is:

a = -9.03 cm / s ^ 2:

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