A particle travels in a straight line(going right) with constant velocity of 4 m/s for 12 seconds. It is then subjected to a constant acceleration in the opposite direction for 20 seconds which returns the particle to its original position. Find the acceleration of the particle.
This is my working:
Left - negative and Right - positive
so the distance travelled right is just (4)(12) since it has constant velocity, a = 0, so it travelled 48 m right.
so next we have the going back journey which is going left back to original position.
a = -a, x = -48 and t = 20 u = 0, since the particle is instantaneously at rest when its turning around
using the formula
 + \frac{1}{2}(-a)(20)^2)

However the answer is -0.64 m/s^2. Can anyone see what i did wrong?