Figure depicts the motion of a particle moving along an $x$ axis with a constant acceleration. What are the magnitude and direction of the particle’s acceleration ?

From the figure, we see that x₀ = –2.0 m, we can apply

$x-x_0=v_{0}t+\frac{1}{2}a{t}^2$

with t = 1.0 s, and then again with t = 2.0 s. This yields two equations for the two unknowns, v₀ and a:

$0.0-\left(-2.0m\right)=v_0\left(1.0s\right)+\frac{1}{2}a{\left(1.0s\right)}^2$

$6.0m-\left(-2.0m\right)=v_0\left(2.0s\right)+\frac{1}{2}a{\left(2.0s\right)}^2$

Solving these simultaneous equations yields the results $v_0=0$and $a=4.0m/s^2$.

The fact that the answer is positive tells us that the acceleration vector points in the +x direction.