Acceleration

Example 2.2 Calculating Displacement: A Subway Train

What are the magnitude and sign of displacements for the motions of the subway train shown in parts (a) and (b) of Figure 2.18?

Strategy

A drawing with a coordinate system is already provided, so we don't need to make a sketch, but we should analyze it to make sure we understand what it is showing. Pay particular attention to the coordinate system. To find displacement, we use the equation $\Delta x=x_{\mathrm{f}}-x_{0}$. This is straightforward since the initial and final positions are given.

Solution

1. Identify the knowns. In the figure we see that $x_{\mathrm{f}}=6.70 \mathrm{~km}$ and $x_{0}=4.70 \mathrm{~km}$ for part (a), and $x_{\mathrm{f}}^{\prime}=3.75 \mathrm{~km}$ and $x_{0}^{\prime}=5.25 \mathrm{~km}$ for part (b).

2. Solve for displacement in part (a).

   $\Delta x=x_{\mathrm{f}}-x_{0}=6.70 \mathrm{~km}-4.70 \mathrm{~km}=+2.00 \mathrm{~km}$

3. Solve for displacement in part (b).

   $\Delta x^{\prime}=x_{\mathrm{f}}^{\prime}-x_{0}^{\prime}=3.75 \mathrm{~km}-5.25 \mathrm{~km}=-1.50 \mathrm{~km}$

Discussion

The direction of the motion in (a) is to the right and therefore its displacement has a positive sign, whereas motion in (b) is to the left and thus has a negative sign.