A cart is given a slight push along a near frictionless track (as shown in the video below).

After the push, the cart is observed to move with a near constant velocity $\vec{v}_{cart} =\langle 1.2, 0, 0 \rangle \dfrac{m}{s}$. Determine its location after 3 seconds.

You need to predict the location of the cart using the information provided and any information that you can collect or assume.

• The cart moves to the right.
• The cart's velocity is given by $\vec{v}_{cart} =\langle 1.2, 0, 0 \rangle \dfrac{m}{s}$.

• The initial location of the cart is not known.

• The interactions of the cart with its surroundings, over the interval that you care about, are negligible. That is, the cart moves with a constant velocity.
• As a result, the average and instantaneous velocity are equivalent.
• We will assume the initial location of the cart is $\vec{r}_{i}$.

• The location of the cart can be predicted using the position update formula, $\vec{r}_f = \vec{r}_i + \vec{v}_{avg} \Delta t$
• The motion of the cart is represented using the following motion diagram.

We can compute the final location,

$$\vec{r}_f = \vec{r}_i + \vec{v}_{avg} \Delta t = \vec{r}_i + \vec{v}_{cart} \Delta t = \vec{r}_i + \langle 1.2, 0, 0 \rangle \dfrac{m}{s} (3 s) = \vec{r}_i + \langle 3.6, 0, 0 \rangle m$$

You might use the video to define an origin such that the initial position of the cart is $\vec{r}_i = \langle 0.4, 1.1, 0 \rangle m$. With that new information, the final location of the cart can be computed exactly,

$$\vec{r}_f = \vec{r}_i + \langle 3.6, 0, 0 \rangle m = \langle 0.4, 1.1, 0 \rangle m + \langle 3.6, 0, 0 \rangle m = \langle 4.0, 1.1, 0 \rangle m$$.

Notice that $y$-position of the cart remained unchanged because all the motion of the cart was in the $x$-direction.