# More Examples on Subway Train Motion

One-dimensional motion of a subway train considered in the examples below. Here we have chosen the x-axis so that + means to the right and − means to the left for displacements, velocities, and accelerations. (a) The subway train moves to the right from x0 to xf. Its displacement Δx is +2.0 km. (b) The train moves to the left from x′0 to x′f. Its displacement Δx′ is −1.5 km. (Note that the prime symbol (′) is used simply to distinguish between displacement in the two different situations. The distances of travel and the size of the cars are on different scales to fit everything into the diagram.) Image credit: OpenStax, College Physics

## Example 1 on Calculating Acceleration: A Subway Train Speeding Up

Suppose the train in the (a) part of the figure above accelerates from rest to 30.0 km/h in the first 20.0 s of its motion. What is its average acceleration during that time interval?

### Strategy

It is worth it at this point to make a simple sketch:

Image credit: OpenStax Physics

This problem involves three steps. First we must determine the change in velocity, then we must determine the change in time, and finally we use these values to calculate the acceleration.

### Solution

1. Identify the knowns. $$v_0 = 0$$ (the trains starts at rest), $$v_{\text{f}} = 30.0\text{ km/h},$$ and $$\Delta t = 20.0\text{ s}.$$

2. Calculate $$\Delta v.$$ Since the train starts from rest, its change in velocity is $$\Delta v = +30.0\text{km/h},$$ where the plus sign means velocity to the right.

3. Plug in known values and solve for the unknown, $$\bar{a}.$$

$$\bar{a} = \cfrac{\Delta v}{\Delta t} = \cfrac{+30.0\text{ km/h}}{20.0\text{ s}}$$

4. Since the units are mixed (we have both hours and seconds for time), we need to convert everything into SI units of meters and seconds. (See the lesson on unit conversion or more guidance.)

$$\bar{a} = \left ( \cfrac{+30\text{ km/h}}{20.0\text{ s}} \right ) \left ( \cfrac{10^3\text{ m}}{1\text{ km}} \right ) \left ( \cfrac{1\text{ h}}{3600\text{ s}} \right ) = 0.417\mathrm{\; m/s^2}$$

### Discussion

The plus sign means that acceleration is to the right. This is reasonable because the train starts from rest and ends up with a velocity to the right (also positive). So acceleration is in the same direction as the change in velocity, as is always the case.

## Example 2 on Calculating Acceleration: A Subway Train Slowing Down

Now suppose that at the end of its trip, the train in the (a) part of the figure above slows to a stop from a speed of 30.0 km/h in 8.00 s. What is its average acceleration while stopping?

### Strategy

Image credit: OpenStax Physics

In this case, the train is decelerating and its acceleration is negative because it is toward the left. As in the previous example, we must find the change in velocity and the change in time and then solve for acceleration.

### Solution

1. Identify the knowns. $$v_0 = 30.0\text{ km/h}, v_{\text{f}} = 0\text{ km/h}$$ (the train is stopped, so its velocity is 0), and $$\Delta t = 8.00\text{ s}.$$

2. Solve for the change in velocity, $$\Delta v.$$

$$\Delta v = v_{\text{f}} \; – \; v_0 = 0 \; – \; 30.0\text{ km/h} = -30.0\text{ km/h}$$

3. Plug in the knowns, $$\Delta v$$ and $$\Delta t,$$ and solve for $$\bar{a}.$$

$$\bar{a} = \cfrac{\Delta v}{\Delta t} = \cfrac{-30.0\text{ km/h}}{8.00\text{ s}}$$

4. Convert the units to meters and seconds.

$$\bar{a} = \cfrac{\Delta v}{\Delta t} = \left ( \cfrac{-30\text{ km/h}}{8.00\text{ s}} \right ) \left ( \cfrac{10^3\text{ m}}{1\text{ km}} \right ) \left ( \cfrac{1\text{ h}}{3600\text{ s}} \right ) = -1.04\mathrm{\; m/s^2}$$

### Discussion

The minus sign indicates that acceleration is to the left. This sign is reasonable because the train initially has a positive velocity in this problem, and a negative acceleration would oppose the motion. Again, acceleration is in the same direction as the change in velocity, which is negative here. This acceleration can be called a deceleration because it has a direction opposite to the velocity.

The graphs of position, velocity, and acceleration vs. time for the trains in example 1 and example 2 above are displayed in the figure below (We have taken the velocity to remain constant from 20 to 40 s, after which the train decelerates.)

(a) Position of the train over time. Notice that the train’s position changes slowly at the beginning of the journey, then more and more quickly as it picks up speed. Its position then changes more slowly as it slows down at the end of the journey. In the middle of the journey, while the velocity remains constant, the position changes at a constant rate. (b) Velocity of the train over time. The train’s velocity increases as it accelerates at the beginning of the journey. It remains the same in the middle of the journey (where there is no acceleration). It decreases as the train decelerates at the end of the journey. (c) The acceleration of the train over time. The train has positive acceleration as it speeds up at the beginning of the journey. It has no acceleration as it travels at constant velocity in the middle of the journey. Its acceleration is negative as it slows down at the end of the journey. Image credit: OpenStax Physics

This is a lesson from the tutorial, One-Dimensional Kinematics and you are encouraged to log in or register, so that you can track your progress.