FHSST Physics/Rectilinear Motion/Graphs

= Graphs =

In physics we often use graphs as important tools for picturing certain concepts. Below are some graphs that help us picture the concepts of displacement, velocity and acceleration.

Displacement-Time Graphs
Below is a graph showing the displacement of the cyclist from A to C:



This graphs shows us how, in 10 seconds time, the cyclist has moved from A to C. We know the gradient (slope) of a graph is defined as the change in y divided by the change in x, i.e. $$\frac{\Delta y}{\Delta x}$$. In this graph the gradient of the graph is just $$\frac{\Delta \overrightarrow{s}}{\Delta t}$$ - and this is just the expression for velocity.

The slope is the same all the way from A to C, so the cyclist's velocity is constant over the entire displacement he travels. In figure 5.1 are examples of the displacement-time graphs you will encounter.



a) shows the graph for an object stationary over a period of time. The gradient is zero, so the object has zero velocity.

b) shows the graph for an object moving at a constant velocity. You can see that the displacement is increasing as time goes on. The gradient, however, stays constant (remember: its the slope of a straight line), so the velocity is constant. Here the gradient is positive, so the object is moving in the direction we have defined as positive.

c) shows the graph for an object moving at a constant acceleration. You can see that both the displacement and the velocity (gradient of the graph) increase with time. The gradient is increasing with time, thus the velocity is increasing with time and the object is accelerating.

Uniform acceleration and gradient (slope)
Look at the velocity-time graph below:



This is the velocity-time graph of a cyclist traveling from A to B at a constant acceleration, i.e. with steadily increasing velocity. The gradient (slope) of this graph is just $$\frac{\Delta \overrightarrow{v}}{\Delta t}$$ - and this is just the expression for acceleration. Because the slope is the same at all points on this graph, the acceleration of the cyclist is constant. And the constant acceleration is $$2 / m/s^2$$, or 2 meters per second per second (or 2 meters per second squared)

$$\begin{matrix}\mathrm{slope \ of \ line} &=& \Delta v / \Delta t \\&=& \frac{10 m/s}{5 s} \\&=& 2 \frac{m}{s^2} \\\end{matrix}$$

Distance travelled
Not only can we get the acceleration of an object from its velocity-time graph, but we can also get some idea of the displacement traveled. Look at the graph below:



This graph shows an object moving at a constant velocity of 10m/s for a duration of 5s. The area between the graph and the time axis (the shaded area) of the above plot will give us the displacement of the object during this time. In this case we just need to calculate the area of a rectangle with width 5s and height 10 m/s

$$\begin{matrix}\mathrm{area \ of \ rectangle} &=& \mathrm{height} \times \mathrm{width}\\&=& \overrightarrow{v} \times t\\&=& 10 \frac{m}{s} \times 5 s\\&=& 50 m\\&=& \overrightarrow{s}= \mathrm{displacement}\\\end{matrix}$$

So, here we've shown that an object traveling at 10 m/s for 5s has undergone a displacement of 50m.

Here are a couple more velocity-time graphs to get used to:

In figure 5.2 are examples of the displacement-time graphs you may encounter.

a) shows the graph for an object moving at a constant velocity over a period of time. The gradient is zero, so the object is not accelerating.

b) shows the graph for an object which is decelerating. You can see that the velocity is decreasing with time. The gradient, however, stays constant (remember: its the slope of a straight line), so the acceleration is constant. Here the gradient is negative, so the object is accelerating in the opposite direction to its motion, hence it is decelerating.

Acceleration-Time Graphs
In this chapter on rectilinear motion we will only deal with objects moving at a constant acceleration, thus all acceleration-time graphs will look like these two:



Here is a description of the graphs below:

a) shows the graph for an object which is either stationary or traveling at a constant velocity. Either way, the acceleration is zero over time.

b) shows the graph for an object moving at a constant acceleration. In this case the acceleration is positive - remember that it can also be negative.

We can obtain the velocity of a particle at some given time from an acceleration time graph - it is just given by the area between the graph and the time-axis. In the graph below, showing an object at a constant positive acceleration, the increase in velocity of the object after 2 seconds corresponds to the shaded portion.



$$\begin{matrix}\mathrm{area \ of \ rectangle} &=& \overrightarrow{a} \times t \\&=& 5 \frac{m}{s^2} \times 2s\\&=& 10 \frac{m}{s}\\&=& \overrightarrow{v}\\\end{matrix}$$

Its useful to remember the set of graphs below when working on problems. Figure 5.3 shows how displacement, velocity and time relate to each other. Given a displacement-time graph like the one on the left, we can plot the corresponding velocity-time graph by remembering that the slope of a displacement-time graph gives the velocity. Similarly, we can plot an acceleration-time graph from the gradient of the velocity-time graph.

Worked Example 24 Relating displacement-, velocity-, and acceleration-time graphs
Question: Given the displacement-time graph below, draw the corresponding velocity-time and acceleration-time graphs, and then describe the motion of the object.



Answer:

Step 1 : Analyse the question to determine what is given. The question explicitly gives a displacement-time graph.

Step 2 : What is asked?

3 things:


 * 1) Draw a velocity-time graph
 * 2) Draw an acceleration-time graph
 * 3) Describe the behaviour of the object

For the first 2 seconds we can see that the displacement remains constant - so the object is not moving, thus it has zero velocity during this time. We can reach this conclusion by another path too: remember that the gradient of a displacement-time graph is the velocity. For the first 2 seconds we can see that the displacement-time graph is a horizontal line, i.e. it has a gradient of zero. Thus the velocity during this time is zero and the object is stationary.

For the next 2 seconds, displacement is increasing with time so the object is moving. Looking at the gradient of the displacement graph we can see that it is not constant. In fact, the slope is getting steeper (the gradient is increasing) as time goes on. Thus, remembering that the gradient of a displacement-time graph is the velocity, the velocity must be increasing with time during this phase.

For the final 2 seconds we see that displacement is still increasing with time, but this time the gradient is constant, so we know that the object is now travelling at a constant velocity, thus the velocity-time graph will be a horizontal line during this stage.

So our velocity-time graph looks like this one below. Because we haven't been given any values on the vertical axis of the displacement-time graph, we cannot figure out what the exact gradients are and hence what the values of the velocity are. In this type of question it is just important to show whether velocities are positive or negative, increasing, decreasing or constant.



Once we have the velocity-time graph its much easier to get the acceleration-time graph as we know that the gradient of a velocity-time graph is the just the acceleration.

For the first 2 seconds the velocity-time graph is horizontal at zero, thus it has a gradient of zero and there is no acceleration during this time. (This makes sense because we know from the displacement time graph that the object is stationary during this time, so it can't be accelerating).

For the next 2 seconds the velocity-time graph has a positive gradient. This gradient is not changing (i.e. its constant) throughout these 2 seconds so there must be a constant positive acceleration.

For the final 2 seconds the object is traveling with a constant velocity. During this time the gradient of the velocity-time graph is once again zero, and thus the object is not accelerating.

The acceleration-time graph looks like this:



A brief description of the motion of the object could read something like this: At t = 0s and object is stationary at some position and remains stationary until t = 2s when it begins accelerating. It accelerates in a positive direction for 2 seconds until t = 4s and then travels at a constant velocity for a further 2 seconds.



Worked Example 25 Calculating distance from a velocity-time graph
Question: The velocity-time graph of a car is plotted below. Calculate the displacement of the car has after 15 seconds.



Answer: We are asked to calculate the displacement of the car. All we need to remember here is that the area between the velocity-time graph and the time axis gives us the displacement.

For t = 0s to t = 5s this is the triangle on the left:

$$\begin{matrix}Area \triangle &=& \frac{1}{2}b \times h\\&=& \frac{1}{2}5s\times 4m/s\\&=&10m\end{matrix}$$

For t = 5s to t = 12s the displacement is equal to the area of the rectangle

$$\begin{matrix}Area \Box &=& w \times h\\&=&7s \times 4m/s\\&=&28m\end{matrix}$$

For t = 12s to t = 14s the displacement is equal to the area of the triangle above the time axis on the right

$$\begin{matrix}Area \triangle &=& \frac{1}{2}b \times h\\&=& \frac{1}{2}2s \times 4m/s\\&=&4m\end{matrix}$$

For t = 14s to t = 15s the displacement is equal to the area of the triangle below the time axis

$$\begin{matrix}Area \triangle &=& \frac{1}{2}b \times h\\&=& \frac{1}{2}1s \times 2m/s\\&=&1m\end{matrix}$$

Now the total displacement of the car is just the sum of all of these areas. HOWEVER, because in the last second (from t = 14s to t = 15s) the velocity of the car is negative, it means that the car was going in the opposite direction, i.e. back where it came from! So, to get the total displacement, we have to add the first 3 areas (those with positive displacements) and subtract the last one (because it signifies a displacement in the opposite direction).

$$\begin{matrix}\overrightarrow{s}&=&10 +28 +4 -1\\&=&41m\ in\ the\ positive\ direction\end{matrix}$$



Worked Example 26 Velocity from a displacement-time graph
Question: Given the diplacement-time graph below,


 * 1) what is the velocity of the object during the first 4 seconds?
 * 2) what is the velocity of the object from t = 4s to t = 7s?



Answer:
 * 1) for the first 4sec velocity is slope of the curve i.e. 2/4=0.5 m/s


 * 1) For the last 3 seconds we can see that the displacement stays constant, and that the gradient is zero. Thus $$\overrightarrow{v}=0m/s$$.



Worked Example 27 From an acceleration- to a velocity-time graph
Question: Given the acceleration-time graph below, assume that the object starts from rest and draw its velocity-time graph.

Answer: