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

The speed was not exceeded so it was slow