Since the cart is released from rest at a distance $x = 0.10text{ m}$ from the equilibrium we conclude that the amplitude of the ensuing oscillating motion is $A = x = 0.10text{ m}$.

The amplitude of the cart’s motion is $A = 0.10text{ m}$

When the position of the cart is $x = – 0.10text{ m}$ it is in its other amplitude.

Remember that the cart started its motion from rest at a distance $x = 0.10text{ m}$.

Since the cart is in its amplitude its speed is 0.

The cart’s speed in $x = – 0.10text{ m}$ is zero since that is the second amplitude.

We know that the amplitude of our motion is $A = 0.25text{ m}$ and the period of our motion is $T = 1.2text{ s}$.

Our body moves during a time interval of $t = 2.4text{ s}$.

Notice first that $t = 2T$, meaning that two full periods of motion will be completed during the time $t$.

We also know from the definition of a single oscillation that the body returns to the initial position after time $T$, having traveled a length of $4A$.

Since two periods are completed the total traveled length will be:

$$
begin{align*}
X = 2 cdot 4A = 8A = 8 cdot 0.25text{ m} = 2text{ m}
end{align*}
$$

The mass will move through a total distance $X = 2text{ m}$