TOPIC 1A: FORCES and MOTION

1.2b Motion graphs

In this section, you will learn how graphs are used to describe the movement of any object. For this course there are two main types of graph used:

- Distance - Time graphs
- Velocity - Time graphs

The time measurement on all these graphs is always given on the bottom axis - the x-axis.

As the name implies, this is simply a graph of the distance covered by any moving object, from a starting point at a time of zero. Let's have a look at a few basic rules, by looking at a few examples:

This graph shows a body that is stationary. The distance is not changing. It is standing still at the start point. |
This graph also shows a body that is stationary. However it is standing still some distance from the start point. |

This graph shows a body moving away from the start at a constant velocity. The steeper the line, the faster the velocity. (The dotted line shows a body moving slower than the solid line). |
This curved line shows a body accelerating. The line is clearly getting steeper (increasing gradient) so the velocity is increasing. |

**Figure 1. Basic Distance-Time Graph Features**

Using these graphs, we can determine the movement of a body, and also by taking readings from the graph we can use the maths from the last section to find the velocity. Remember that **velocity** basically means the same as **speed** in these graphs. have a go at these questions to check you understand this section:

**Questions:**

1. A man is taking his dog for a walk. When he opens the front door, the dog runs off in a straight line at a steady speed. It then stops at a lamp post for some time.

Sketch a distance-time graph for this motion. You do not need to estimate any values for this question.

- The distance should be labelled on the y-axis.
- The time should be labelled on the x-axis.
- The first section of the line should be a straight line sloping upwards showing "a steady speed".
- The next section should be a flat, horizontal line showing the dog remaining in one spot at the lamp post.

Here is an example:

2. A student writes a program to make a small robot move across the desk. This graph shows the motion of the robot:

- a). Describe the motion in part 'X' on the graph.
- b). Describe the motion in part 'Y' on the graph.
- c). Describe how the motion at 'Z' is different from at 'X'.
- d) Calculate the speed of the robot when it is moving at its fastest.

a) Line 'X' shows the robot moving away from the start at **constant velocity**.

b) Line 'Y' shows the robot **stops** / is stationary (100 cm from the start).

c) Line 'Z' shows the robot moving at constant velocity just as line 'X', but this time at a slower velocity as the line is less steep. (Lower gradient).

d) The fastest speed is where the line is the steepest, (the highest gradient), which is line 'X'. For this section, reading from the graph, the robot covers 100 cm in 5 seconds. As we are using cm, the speed will be given in cm/s.

speed = | distance | = | 100 cm |

time | 5 s |

so the speed, or velocity, is = **20 cm /s.**

( We could convert the distance measurement to metres to give a velocity of 0.2 m/s)

These graphs are more frequently used in exams, as they are slightly harder and give more information about the motion of the body. They are often called **speed- time graphs**. Again, it is simply a plot of the velocity/speed on the y-axis against time on the x-axis.

Here are the 4 most important features you need to learn on these graphs:

This graph shows a body that is stationary. The velocity is zero so it is not moving. |
This graph shows a body that is moving with a constant velocity. |

This graph shows a body that is speeding up, or accelerating. The steeper the line, the higher the acceleration. (The solid line shows a higher acceleration than the the dotted line). |
This graph shows a body that is slowing down, or decelerating. (You can also describe this a negative acceleration). |

**Figure 2. Basic Velocity-Time Graph Features**

As you can see, these graphs look very similar to the distance-time graphs and they are easily confused. Take care to check the label on the y-axis to make sure you know which is which. Some people just seem to have a knack for understanding what any motion graph is showing. If you find this more difficult, you need to** learn these shapes.**

As shown above, deceleration can be described as a negative acceleration. So if a car is braking, we can say that is decelerating at (for example) 2 m/s^{2, }or say that it is accelerating at -2 m/s^{2}.

A velocity-time graph is often used to show motion because you can use it to find other information. The two that we will cover here are how to find the **acceleration** and the **distance travelled** from the graph.

**The distance travelled = the area under the line.**

**The acceleration = the gradient of the line.**

Take this example, showing a cyclist accelerating way from traffic lights:

**Figure 3. An Accelerating Cyclist**

The line in this graph slopes upwards, showing acceleration. We can use the information to calculate the distance travelled in this time and also the acceleration:

**Distance:** This is the area under the line (between the red line and the x-axis), shown by the shaded area in this example. The area of this triangular shape is:

Area = ½ x base x height = ½ x 30 x 6

So the **distance travelled = 90 m**.

(The units should be 'm' because the velocity is given in 'm/s').

**Acceleration:** This is the gradient of the line. figure 2 above shows that the steeper the line, the higher the acceleration. The 'steepness' of a line is the gradient. From your maths lessons, you should know that:

gradient = | 'rise' | or gradient = | change in y-axis |

'run' | change in x-axis |

On the example above shown in figure 3, the change in the y-axis is 6 m/s, and the change in the x-axis is 30 s. Therefore:

gradient = | 6 | = 0.2 |

30 |

So the **acceleration = 0.2 m/s ^{2}**.

This section covers the maths behind the meaning of the gradient and area under the line on a velocity-time graph.

**1. Acceleration:**

The formula for acceleration is:

acceleration = | change in velocity |

time taken |

However on a velocity time graph the change in velocity is the change in the y-axis, and the time taken is the change in the x-axis. Therefore:

acceleration = |
change in velocity | = | change in y-axis | = gradient |

time taken | change in x-axis |

**2. Distance:**

Have a look at this graph:

We know that distance travelled = speed x time.

However in this graph:

speed = velocity = height of the grey shaded box;

time = length of grey shaded box.

Therefore (distance = speed x time) becomes (distance = height x width) which is the area of the box!

**So distance travelled = area under the line.**

This is true for all shapes - just find the area under the line showing the velocity of the object.

Now have a go at some questions about velocity-time graphs:

**Questions:**

The following graph is for use with **questions 3 & 4, **and shows the motion of a tram:

**Figure 4. A Velocity time graph for a tram.**

3. The graph shown in figure 4 gives data on the movement of a city tram moving away from a station.

Which section or sections of the graph shows the tram ...

- a) moving with constant velocity?
- b) decelerating?
- c) accelerating at the highest rate?

a) A constant velocity will be shown as a flat, horizontal line on a velocity-time graph.

Therefore the**sections showing this are Q and S.**

b) Deceleration means slowing down, and this is shown by a sloping line with a decreasing velocity.

The section showing this is**section T.**

c) A high acceleration is shown by a sloping line with a high gradient (like a very steep hill).

This is shown by**section R. **

(Section P is also showing acceleration, but the tram is not accelerating as much as in section R).

Therefore the

The section showing this is

This is shown by

(Section P is also showing acceleration, but the tram is not accelerating as much as in section R).

4. Using the graph shown in figure 4 for the velocity of a tram, calculate:

- a) The distance travelled between 30 and 40 seconds.
- b) The distance travelled in the first 10 seconds.
- c) The acceleration of the tram in section R.

a) The distance travelled between 30 and 40 s is given by:

Distance = area under line = rectangle of height 15 (m/s) and length 10 (s).

Therefore the **area = distance = 15 x 10 = 150 m. **

b) The distance travelled in the first 10 s is given by:

Distance = area under line = area of triangle = ½ base x height,

So area =
½ x 10 x 5,

Area = **distance = 25 m. **

c) The acceleration is given by:

acceleration = |
change in velocity |

time taken |

or by using acceleration = gradient = |
change in y-axis |

change in x-axis |

Both methods give:

acceleration = |
15 - 5 | = | 10 |

10 | 10 |

So the **acceleration of section R = 1 m/s ^{2}.**

A cyclist takes part in a sprint event on a track. At the start of the race, the cyclist produces as large a force as possible to make the bike accelerate. However as he goes faster and faster, air resistance and friction push backwards on the cyclist and bike.

Have a look at the diagram below to see exactly what is happening. In this example, we have ignored friction as a negligible force, and only included the force from the cyclist and air resistance:

**Figure 5. A cyclist reaching terminal velocity**

Let's look at each stage of figure 5 in a bit more detail:

**Diagram A:**At the beginning of the race, the cyclist starts at a speed of 0 m/s . There is no air resistance at this speed, just a forward force produced by the cyclist. The cyclist accelerates.**Diagram B:**The cyclist has reached 6 m/s. At this speed, the air resistance is significant. What will happen now?

As the forward force F_{c}is still the same, and larger than the air resistance, there will be a resultant forward force. This will continue to make the cyclist accelerate, but at a lower rate than at the start.**Diagram C:**Here the cyclist is now going so fast that the air resistance is equal to the forward force F_{c}- the two forces are balanced. The cyclist does not stop suddenly -(a common answer!) - he will stay at a constant velocity. Neither force 'wins' and the net result is the velocity stays constant. This 'top speed' of the cyclist is called the**terminal velocity.**

All of this can be summarised in a velocity - time graph:

**Figure 6. A terminal velocity graph**

This is a very famous graph and is often used in the exams - you will need to learn how to draw it and explain each part of the process of reaching terminal velocity. Other common examples include skydivers, and other falling objects like squash balls - all reach a different terminal velocity depending on how air resistance affects them. For a falling object, the downward force is the weight (the force of gravity), and air resistance acts upwards.

Remember that liquids also add a resistive force, so this graph could be used for a ball falling through water or **any number of experiments where objects move through a gas or a liquid. **

**Questions:**

5. The diagrams below (figure 4) show forces acting on a skydiver who has jumped out of a plane. There are 3 diagrams, showing different stages of the descent.

**Figure 7. The forces acting on a skydiver**

- a). Describe the acceleration in diagram A.
- b). Describe the acceleration in diagram B.
- c). Describe the acceleration in diagram C.
- d). State the causes of the forces acting downwards and upwards on the skydiver.
- e). Complete the velocity time graph shown below to show each of these stages of the skydiver's fall from figure 4. You do not need to include numbers on the graph.

The initial downwards velocity of the skydiver is zero, as shown by the line section given at the start of the graph.

- a). The downwards force in diagram A is larger (by 300 N) and so the skydiver
**accelerates downwards.** - b). The forces are balanced, so the skydiver stays at a constant velocity. Remember the question asks you to describe acceleration - the
**acceleration is zero**. - c). There is now a resultant force upwards. Be careful - this does NOT make the skydiver go back up to the plane!. They are falling downwards, so the unbalanced force just slows them down. The
**skydiver decelerates**(or has negative acceleration). - d) The downwards force is the
**weight (force of gravity)**, and the upwards force is**air resistance**. - e) See the example here:

When any object falls under gravity on Earth, it always initially accelerates downwards at 9.8 m/s^{2}. This is called the acceleration of gravity on Earth (given the letter 'g'), and is independent of the mass of the object.

You can drop a heavy ball or a feather, and IF there is no air resistance, they will both accelerate downwards at g= 9.8 m/s^{2} as shown in this video. Note that the final fall in the video is in slow motion!

**YouTube video - The world's largest vacuum chamber (Human Universe - BBC)**

**Now test your understanding using these quick, 10 minute questions on this topic from Grade Gorilla:**