A subway train accelerates uniformly from rest to [latex]\scriptsize 30.0\ \text{km/h}[/latex] in the first [latex]\scriptsize 20\ \text{s}[/latex] of its journey between stops.

1. Draw an accurate velocity–time graph of this situation.
2. Use the graph to determine the uniform rate of acceleration.
3. Draw an accurate acceleration–time graph of this situation.
4. Draw an approximate displacement–time graph of this situation.

Solutions

1. The first thing we should do is convert the velocity into standard units.
   [latex]\scriptsize 30.0\ \text{km/h}=8.33\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}[/latex]
   .
   We are told that the train accelerates uniformly i.e. at the same rate throughout the period. Therefore, we know that the gradient of the velocity–time graph will be constant. In other words, it will be a straight line. The train also starts from rest and reaches a velocity of [latex]\scriptsize 8.33\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}[/latex] after [latex]\scriptsize 20\ \text{s}[/latex]. Therefore, the velocity–time graph will look as follows:
   
2. To find the rate of acceleration, we have to measure the gradient of the velocity–time graph. Because the acceleration is uniform, we know the gradient of the line is the same for the whole line. Therefore, we can measure the gradient between any two convenient points, even the points at the start and end of the line.
   .
   [latex]\scriptsize m=a=\displaystyle \frac{{{{v}_{f}}-{{v}_{i}}}}{{{{t}_{f}}-{{t}_{i}}}}=\displaystyle \frac{{8.33\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}-0\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}}}{{20\ \text{s}-0\ \text{s}}}=\displaystyle \frac{{8.33\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}}}{{20\ \text{s}}}=0.427\ \text{m}\text{.}{{\text{s}}^{{\text{-2}}}}[/latex]
3. We know that the acceleration is a constant [latex]\scriptsize 0.43\ \text{m}\text{.}{{\text{s}}^{{\text{-2}}}}[/latex].
   
4. For a body that is accelerating at a constant rate, its velocity is increasing as time passes and so its displacement is increasing at an increasing rate. In other words, the gradient of the displacement–time graph is not a straight line because a changing gradient reflects the changing velocity. As the velocity increases, the gradient of the displacement–time graph must also increase.
   .
   We have not added values to the y-axis because we do not know precisely what these are. We could use the equations of motion to calculate the maximum displacement of the train at [latex]\scriptsize 20\ \text{s}[/latex].
   

Suppose the subway train from example 3.1 slows uniformly to a stop at the next station from a speed of [latex]\scriptsize 30.0\ \text{km/h}[/latex] in [latex]\scriptsize 8\ \text{s}[/latex].

1. Draw an accurate velocity–time graph of this situation, taking the time when it starts to slow down as [latex]\scriptsize t=0\ \text{s}[/latex].
2. Use the graph to determine the uniform rate of acceleration.
3. Draw an accurate acceleration–time graph of this situation.
4. Draw an approximate displacement–time graph of this situation.
5. If between accelerating to [latex]\scriptsize 30.0\ \text{km/h}[/latex] and then decelerating to a stop again, the train travels at a constant velocity for [latex]\scriptsize 20\ \text{s}[/latex], draw velocity–time, acceleration–time and displacement–time graphs of the train’s entire journey from the first stop to the second.

Solutions

1. In this case, the train’s velocity will decrease at a constant rate. Therefore, its gradient will be constant but negative. We know from example 3.1 that its initial velocity is [latex]\scriptsize 8.33\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}[/latex].
   
2. To find the rate of acceleration, we have to measure the gradient of the velocity–time graph. Because the acceleration is uniform, we know the gradient of the line is the same for the whole line. Therefore, we can measure the gradient between any two convenient points, even the points at the start and end of the line. We can already see that, because the velocity–time graph gradient is negative, our acceleration will be negative (indicating deceleration).
   .
   [latex]\scriptsize m=a=\displaystyle \frac{{{{v}_{f}}-{{v}_{i}}}}{{{{t}_{f}}-{{t}_{i}}}}=\displaystyle \frac{{0\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}-8.33\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}}}{{8\ \text{s}-0\ \text{s}}}=\displaystyle \frac{{-8.33\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}}}{{8\ \text{s}}}=-1.04\ \text{m}\text{.}{{\text{s}}^{{\text{-2}}}}[/latex]
3. We know that the acceleration is a constant [latex]\scriptsize -1.04\ \text{m}\text{.}{{\text{s}}^{{\text{-2}}}}[/latex].
   
4. For a body that is decelerating at a constant rate, its velocity is decreasing as time passes and so its displacement is still increasing but at a decreasing rate. In other words, the gradient of the displacement–time graph is not a straight line because a changing gradient reflects the changing velocity. As the velocity decreases, the gradient of the displacement–time graph must also decrease.
   
5. Here is the velocity–time graph:
   
   Between A and B, the train accelerated from [latex]\scriptsize 0\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}[/latex] to [latex]\scriptsize 8.33\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}[/latex]. Between B and C, the train travelled at a constant velocity of [latex]\scriptsize 8.33\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}[/latex]. Between C and D, the train decelerated from [latex]\scriptsize 8.33\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}[/latex] to [latex]\scriptsize 0\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}[/latex].
   .
   Here is the acceleration–time graph:
   
   From A to B, the train accelerated at a constant rate of [latex]\scriptsize 0.43\ \text{m}\text{.}{{\text{s}}^{{\text{-2}}}}[/latex]. Between C and D, the train travelled at a constant velocity, hence its acceleration was [latex]\scriptsize 0\ \text{m}\text{.}{{\text{s}}^{{\text{-2}}}}[/latex]. Between E and F, the train accelerated at a constant rate of [latex]\scriptsize -1.04\ \text{m}\text{.}{{\text{s}}^{{\text{-2}}}}[/latex] (decelerated).
   .
   Here is the displacement–time graph:
   
   Between A and B, the train’s velocity was increasing. Therefore, the train’s displacement was increasing at an increasing rate. The total displacement after [latex]\scriptsize 20\ \text{s}[/latex] can be calculated as:
   [latex]\scriptsize \begin{align*}s&=ut+\displaystyle \frac{1}{2}a{{t}^{2}}\\&=0\times 20\ \text{s}+\displaystyle \frac{1}{2}\times 0.43\ \text{m}\text{.}{{\text{s}}^{{-2}}}\times {{(20\ \text{s})}^{2}}\\&=86\ \text{m}\end{align*}[/latex]
   .
   Between B and C, the train travelled at a constant velocity of [latex]\scriptsize 8.33\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}[/latex]. Therefore, the graph is a straight line with a positive gradient of [latex]\scriptsize 8.33\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}[/latex]. The displacement after these [latex]\scriptsize 20\ \text{s}[/latex] can be calculated as:
   [latex]\scriptsize \begin{align*}s&=ut+\displaystyle \frac{1}{2}a{{t}^{2}}\\&=8.33\ \text{m}\text{.}{{\text{s}}^{{-1}}}\times 20\ \text{s}+\displaystyle \frac{1}{2}\times 0\ \text{m}\text{.}{{\text{s}}^{{-2}}}\times {{(20\ \text{s})}^{2}}\\&=166.6\ \text{m}\end{align*}[/latex]
   .
   So, the train’s total displacement at this point is [latex]\scriptsize 86\ \text{m}+166.6\ \text{m}=252.6\ \text{m}[/latex].
   .
   Between C and D, the train’s velocity decreased and so its displacement increased but at a decreasing rate. The displacement during these [latex]\scriptsize 8\ \text{s}[/latex] can be calculated as:
   [latex]\scriptsize \begin{align*}s&=ut+\displaystyle \frac{1}{2}a{{t}^{2}}\\&=8.33\ \text{m}\text{.}{{\text{s}}^{{-1}}}\times 8\ \text{s}+\displaystyle \frac{1}{2}\times \left( {-1.04\ \text{m}\text{.}{{\text{s}}^{{-2}}}} \right)\times {{(8\ \text{s})}^{2}}\\&=33.36\ \text{m}\end{align*}[/latex]
   .
   Therefore, the train’s total displacement at this point is:
   [latex]\scriptsize 86\ \text{m}+166.6\ \text{m}+33.36\ \text{m}=285.96\ \text{m}[/latex]
   .
   Note: You will not be expected to draw an accurate displacement–time graph for this kind of situation. However, it is important that you understand why the graph is shaped like it is and to see the direct link between the graph and the equations of motion we covered in unit 1.

A subway train approaches a station travelling at a constant velocity of [latex]\scriptsize 45\ \text{km/h}[/latex]. After [latex]\scriptsize 10\ \text{s}[/latex], it decelerates uniformly to rest in [latex]\scriptsize 12\ \text{s}[/latex]. Unfortunately, it overshoots the station and has to reverse, which it does so immediately, accelerating uniformly to [latex]\scriptsize 10\ \text{km/h}[/latex] in [latex]\scriptsize 5\ \text{s}[/latex] and then immediately decelerating to rest at the station in [latex]\scriptsize 5\ \text{s}[/latex].

1. Draw an accurate velocity–time graph of this situation.
2. Draw an accurate acceleration–time graph of this situation.
3. Draw an approximate displacement–time graph of this situation.
4. Use one of your graphs to determine the distance the train had to reverse back to the station.

Solutions

Before we begin answering any of the questions, we need to define the positive direction for all our vectors. We will assign the initial direction the train travelled to the station as the positive direction and the direction it travelled while reversing back to the station as the negative direction.

1. [latex]\scriptsize 45\ \text{km/h}=12.5\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}[/latex]
   [latex]\scriptsize 10\ \text{km/h}=2.78\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}[/latex]
   The velocity–time graph is as follows:
   
   Between A and B, the train is travelling at a constant speed of [latex]\scriptsize 12.5\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}[/latex] for [latex]\scriptsize 10\ \text{s}[/latex]. Between B and C, the train is decelerating uniformly to rest (hence the gradient of the line is constant but negative). From C to D, the train is accelerating to [latex]\scriptsize 2.78\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}[/latex]. However, this is in the negative direction. So, the actual velocity reached after [latex]\scriptsize 5\ \text{s}[/latex] of uniform acceleration is [latex]\scriptsize -2.78\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}[/latex] and the graph is below the x-axis. From D to E the train is decelerating uniformly to rest in [latex]\scriptsize 5\ \text{s}[/latex]. During this period, the train is still travelling in the negative direction and so the graph is below the x-axis.
2. The acceleration–time graph is as follows:
   
   Between A and B, the train is travelling at a constant velocity. Therefore, the acceleration is zero. The gradient of the velocity–time graph during this period is zero. Between B and C, the train is decelerating to rest at [latex]\scriptsize -1.04\ \text{m}\text{.}{{\text{s}}^{{\text{-2}}}}[/latex]. The acceleration for this period is negative and so is drawn below the x-axis. We can see that the gradient of the velocity–time graph is constant and negative during this period. Between C and D the train is accelerating from rest to [latex]\scriptsize 2.78\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}[/latex]. It may seem strange, however, that this part of the graph is also negative. If you look at the velocity-time graph for this period (C to D), you will see that its gradient is a constant but negative [latex]\scriptsize 0.56\ \text{m}\text{.}{{\text{s}}^{{\text{-2}}}}[/latex]. Even though the train is accelerating (which we normally take to be positive acceleration), it is accelerating in the negative direction. Therefore, this acceleration is [latex]\scriptsize -0.56\ \text{m}\text{.}{{\text{s}}^{{\text{-2}}}}[/latex].
   .
   Between D and E, the train is decelerating to rest. This is in the negative direction so we can think of this deceleration as [latex]\scriptsize -\left( {-0.56\ \text{m}\text{.}{{\text{s}}^{{\text{-2}}}}} \right)=0.56\ \text{m}\text{.}{{\text{s}}^{{\text{-2}}}}[/latex] i.e. a positive acceleration. We can also see that during this period, the gradient of the velocity–time graph is constant and positive. Therefore, this part of the acceleration–time graph is above the x-axis.
3. The displacement–time graph has the following shape:
   
   Between A and B, the train is travelling at a constant velocity of [latex]\scriptsize 12.5\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}[/latex]. Therefore, the displacement–time graph has a constant positive gradient of [latex]\scriptsize 12.5\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}[/latex]. Between B and C, the train is decelerating but still travelling in the positive direction. Therefore, its displacement is still increasing but at a decreasing rate as the velocity decreases. Therefore, the displacement–time graph gradient is decreasing. Between C and D, the train starts reversing. Therefore, its displacement is decreasing. Its velocity is negative and increasing and so the gradient of the displacement–time graph is negative and getting more negative. From D to E, the train slows down but is still travelling in the negative direction. Therefore, displacement is still decreasing but at a slower rate as the negative velocity is decreasing and so the gradient of the displacement–time graph is negative but getting less negative.
4. The best graph to use is the velocity–time graph. We need to calculate the area under the graph between points C and E to determine how far the train needed to travel in reverse to get back to the station.
   .
   We need to split this area into two triangles. Both will have a base of [latex]\scriptsize 5\ \text{s}[/latex] and a vertical height of [latex]\scriptsize 2.78\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}[/latex]. Because we are calculating area, we do not need to take the negative nature of the velocity into account.
   .
   [latex]\scriptsize \displaystyle \text{Area}=2\times \displaystyle \frac{1}{2}\times 5\ \text{s}\times 2.78\ \text{m}\text{.}{{\text{s}}^{{\text{-1}}}}=13.9\ \text{m}[/latex]
   Therefore, the train missed the station by [latex]\scriptsize \displaystyle 13.9\ \text{m}[/latex].