An object moving with uniform acceleration has a velocity of 14.0 cm/s in the positive x-direction when its x-coordinate is 2.85 cm. If its x-coordinate 3.15 s later is - 5.00 cm, what is its acceleration?

Though expressed in coordinate form, this is an example of linear motion. With the given, the acceleration can be solved using the following equation:

s = Vo*t + (1/2) * a*t^2

Where:

s = displacement = 7.85 cm

Vo = 14.0 cm/s

t = 3.15 s

a = ?

Substituting:

7.85 = (14) * (3.15) + (1/2) * a * (3.15) ^2

a = - 7.31 cm / s^2