An object moving with uniform acceleration has a velocity of 10.5 cm/s in the positive x-direction when its x-coordinate is 2.72 cm. If its x-coordinate 2.30 s later is ?5.00 cm, what is its acceleration? The object has moved to a particular coordinate in the positive x-direction with a certain velocity and constant acceleration; then it reverses its direction and moves in the negative x-direction to a particular x-coordinate in time t. We are given an initial velocity vi = 10.5 cm/s in the positive x-direction when the initial position is xi = 2.72 cm (t = 0). We are given that at t = 2.30 s, the final position is xf = ?5.00 cm. The acceleration is uniform so that we have the following equation in terms of the constant acceleration a. Xf-Xi=Vit-1/2at^2 Now we substitute the given values into this equation. (___cm)-(___cm)=(___cm/s)(__s)+1/2a(___s)