Lesson Explainer: Speed | Nagwa Lesson Explainer: Speed | Nagwa

Lesson Explainer: Speed Science • Third Year of Preparatory School

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In this explainer, we will learn how to determine the speed of an object that moves a distance in a time.

Consider two runners competing in a 100 m race. We will label them “runner A” and “runner B.”

Let’s suppose that runner A is faster than runner B. This means we know that runner A will win the race.

The fact that runner A won the race means that they finished running the 100 m distance before runner B did. In other words, because they were running faster, runner A took less time than runner B to run the same distance.

We can describe how fast or slow something moves using a quantity called speed. We can say that the faster runner was traveling at a greater speed than the slower runner.

So far, we have considered one specific situation involving two runners. However, what we have learnt applies far more generally. Any object that moves must have a particular speed. This speed is a quantity that relates the distance covered by the object and time taken for the object to cover that distance.

Specifically, we can say the following.

Definition: Speed

The speed of an object is a measure of the distance traveled by the object per a second that it travels.

What this means is that the faster an object is moving, or the greater the speed of that object, the more distance that object will travel in a certain amount of time.

We have seen that a faster object travels a greater distance per second than a slower object. Suppose, however, that we compare two objects that travel equal distances, like the two runners who ran the 100 m race. In this case, the greater the speed of the object, the less time that object will take to travel the distance. For our two runners, runner A had the greater speed and so took less time to complete the 100 m distance of the race than runner B did.

Let’s look at a couple of examples.

Example 1: Identifying Which of Two Cars Travels the Greater Distance in a Given Amount of Time

Two cars drive for the same amount of time at different speeds. Which car travels the greater distance?

  1. The car with greater speed
  2. The car with less speed

Answer

This question tells us that we have two cars that travel for the same amount of time. We are asked to work out which of the two cars travels a greater distance in this time.

Let’s consider two cars that are traveling along a road. We will imagine these cars pass the same point on this road at some time that we will refer to as the “start time.” Then, these cars both continue driving for the same amount of time as each other. At a future point in time, which we will refer to as the “end time,” each car has traveled some distance along the road.

We have illustrated this situation in the diagram below. On the left of the diagram, we have the two cars passing the same location at the start time. To the right, we have these two cars after each has traveled on for the same amount of time.

We are told in the question that the two cars travel at different speeds. This means that we have a faster car and a slower car.

We can recall that the speed of an object is a measure of how far that object travels in a given amount of time. The greater the speed, the further the object travels per unit time.

In this case, both cars travel for the same amount of time. Therefore, we know that, in this time, the car with the greater speed, or the faster car, will travel a greater distance. In the sketch that we drew, this means that the red car is the faster car and the blue car is the slower car.

So, our answer to the question is that the car that travels the greater distance is given by option A: the car with greater speed.

Example 2: Identifying Which of Two Cars Takes More Time to Travel a Given Distance

Two cars drive the same distance at different speeds. Which car takes more time to travel the distance?

  1. The car with greater speed
  2. The car with less speed

Answer

This question tells us that we have two cars that travel the same distance. We are asked to work out which of the two takes more time to do this.

Let’s consider two cars traveling along a road. We will imagine these cars pass the same “start point” on this road at the same time as each other. They then both continue driving the same distance on until each passes a point further along the road that we will call the “end point.”

We have illustrated this in the diagram below. On the left of the diagram, we have the two cars passing the same location—the start point—at the same time. To the right, we have these two cars after each has traveled the same distance along the road, such that they are passing the end point.

The question that we need to answer is this: which car took longer to travel the distance between the start point and the end point?

We are told in the question that the cars travel at different speeds. This means we have a faster car (the car with greater speed) and a slower car (the car with less speed).

We can recall that we can think of speed as telling us how long it will take an object to travel a certain distance. In other words, the greater the speed of an object, the less time that object will take to travel a given distance.

In our case, we have two cars that travel the same distance. Using our understanding of speed, we can see that the car with greater speed will take less time to travel this distance. Therefore, the car with less speed (the slower car) will be the car to take more time to do so.

So, our answer to the question is that the car that takes more time to travel the distance is given by option B: the car with less speed.

If the speed of an object does not change during its motion, then we say that the object moves with a constant speed. Since we know that the speed of an object tells us how much distance the object travels in a given amount of time, then we can define constant speed as follows.

Definition: Constant Speed

An object that travels at a constant speed covers equal distances in equal intervals of time.

Let’s have a look at what this definition really means.

We will consider a runner who sprints for a time of 5 s, and who covers a distance of 50 m in this time.

We can split up the total time into smaller intervals of time. Let’s consider intervals of 1 s. We have five of these intervals in the total time of 5 s.

If the runner runs with a constant speed, this means that they must travel the same distance in each of these five intervals of 1 s. So, we need to split the 50 m distance up into five equal segments.

Dividing 50 m by 5, we see that the runner covers a distance of 10 m in each 1 s time interval.

Diagrams like the one above can provide a useful visual way of comparing two objects moving at different speeds to each other.

Let’s suppose we have a second runner, also sprinting at a constant speed. However, we will suppose that this second runner only covers a distance of 8 m in each 1 s time interval. We will refer to this second runner as “runner B”, and the first runner (who covers 10 m in each 1 s time interval) as “runner A.”

We can draw the motion of the two runners as follows:

We have shown each runner’s motion for the same number of time intervals. We can see that, after these five intervals of 1 s, runner A has covered a greater total distance than runner B.

Therefore, runner A must have been moving faster.

We can also look at this in a different way.

Consider the following diagram.

This shows the same two runners as before, but now we can see that they both cover the same total distance as each other. That total distance is 40 m.

We can see that runner A, who runs a distance of 10 m in each 1 s time interval, runs for four such time intervals in order to cover this distance.

Meanwhile, runner B, who runs a distance of 8 m in each 1 s time interval, runs for five time intervals in order to cover this same distance.

Since runner A runs for less time intervals than runner B, this means that they run for less total time. Both runners cover the same distance as each other. So, since runner A takes less time to do this, we know that runner A must be running faster than runner B.

We can also express speed mathematically. Let’s consider an object moving with a constant speed. We will label the distance traveled by that object as 𝑑, the time taken for it to travel that distance as 𝑡, and the speed of the object as 𝑣.

The speed of the object is the amount of distance it travels per unit time. We could also express this statement as speeddistancetime=.

In terms of symbols, we then get the following equation.

Equation: Speed

If an object moving at a constant speed of 𝑣 travels a distance of 𝑑 in a time of 𝑡, then the speed 𝑣 is given by 𝑣=𝑑𝑡.

Whenever we have a mathematical equation like this one for speed, we need to be careful with the units of the quantities. Specifically, the units of the right-hand side of the equation must agree with the units of the left-hand side.

We often measure distance in units of metres (m) and time in units of seconds (s). Then, the units on the right-hand side of the speed equation are units of metres divided by units of seconds. We write these units as “m/s,” and we read this out loud as “metres per second.

Since the units on both sides of the equation must agree, then our speed must also have units of m/s (metres per second).

Let’s look at an example where we need to use our equation for speed.

Example 3: Calculating the Speed of an Object

A bike moves a distance of 15 metres in a time of 3 seconds. What is the speed of the bike?

Answer

In this question, we have a bike that we are told that it moves a distance of 15 m in a time of 3 s. We are asked to find the bike’s speed.

Let’s begin by making a quick sketch of the information we are given.

As shown in the above picture, the bike starts at some position, and a time of 3 s later, it has moved a distance of 15 m from where it started.

In our sketch, we have labeled the distance traveled as 𝑑, so that we have 𝑑=15m. We have also labeled the time taken as 𝑡, so that we have 𝑡=3s.

Additionally, we will label the speed of the bike as 𝑣.

We can recall the equation to calculate the speed 𝑣 of an object that travels a distance 𝑑 in an amount of time 𝑡: 𝑣=𝑑𝑡.

Substituting in the values for 𝑑 and 𝑡, we get the following expression for the speed 𝑣: 𝑣=153.ms

Since the distance has units of metres (m) and the time has units of seconds (s), we know that our speed must have units of metres per second (m/s).

Doing the division on the right-hand side, we then find that 𝑣=153/=5/.msms

So, the speed of the bike is 5 m/s.

We have an equation that relates the speed of an object, the distance traveled by that object, and the time taken to travel that distance. The form that we have seen this equation in is useful if we know the distance and time, and we want to calculate the speed. This is because the speed, 𝑣, is the subject of the equation; in other words, the equation has “𝑣=” on the left-hand side.

We can also rearrange this equation to make the distance, 𝑑, the subject. This is useful if we have an object moving at a constant speed and we know the value of this speed and the time the object moves for, but we want to calculate the distance it moves.

Similarly, we can rearrange the equation to make the time, 𝑡, the subject. Then, if we know the speed of an object and the distance that it travels, we can use the rearranged equation to calculate the time that the object must have taken to travel this distance.

Let’s see how this works. We will begin by rearranging to make 𝑑 the subject.

We start with our equation in the form we have seen previously: 𝑣=𝑑𝑡.

We want to end up with an equation that says “𝑑=.

The golden rule for rearranging equations is that whatever we do to one side of the equation, we must also do to the other side.

We see that since the 𝑑 on the right-hand side is currently divided by 𝑡, we need to multiply the right-hand side by 𝑡. Since we must also do the same to the left-hand side, this means that we need to multiply both sides of the equation by 𝑡. This gives us 𝑣×𝑡=𝑑×𝑡𝑡.

On the right-hand side of the equation, we now have a 𝑡 in both the numerator and the denominator of the fraction. These two 𝑡s cancel each other out, since 𝑡÷𝑡=1.

Once we have canceled the 𝑡s on the right-hand side, we have 𝑣×𝑡=𝑑.

Now, if 𝑣×𝑡 is equal to 𝑑, then it must also be true that 𝑑 is equal to 𝑣×𝑡. This means that we can also write the equation “𝑣×𝑡=𝑑” as 𝑑=𝑣×𝑡.

These two equations are exactly equivalent statements. They are both telling us that if we multiply together the quantities 𝑣 and 𝑡, the value we get will be the same as the value of 𝑑.

When we have an equation in which we multiply together two quantities represented by symbols, sometimes the multiplication sign gets omitted. When we see two symbols or letters next to each other, the multiplication sign is implied even if it is not written out explicitly. So, we may also see this equation written as 𝑑=𝑣𝑡.

This means exactly the same as the previous equation. That is, that the distance 𝑑 traveled by an object moving at a constant speed 𝑣 for a time 𝑡 is equal to 𝑣 multiplied by 𝑡.

Now, let’s consider how we may rearrange the equation to make the time 𝑡 the subject. We will begin again from the version of the equation that we first met: 𝑣=𝑑𝑡.

Since we want to make 𝑡 the subject, a good first step will be to get it out of the denominator of the fraction on the right-hand side. So, as in the previous rearrangement, we begin by multiplying both sides of the equation by 𝑡. This gives us 𝑣×𝑡=𝑑×𝑡𝑡.

Again, as before, the 𝑡s in the numerator and denominator of the fraction of the right-hand side cancel each other out. This gives us 𝑣×𝑡=𝑑.

This time, we want to make 𝑡 the subject; that is, we want an equation that says “𝑡=.” Currently, we have 𝑡 on the left-hand side of the equation, but it is multiplied by 𝑣. So, we need to divide both sides of the equation by 𝑣: 𝑣×𝑡𝑣=𝑑𝑣.

On the left-hand side of the equation, we have a 𝑣 in the numerator and the denominator of the fraction. These two 𝑣s cancel each other out, leaving us with 𝑡=𝑑𝑣.

This equation tells us how to calculate the time that an object travels for, given the speed of that object and the distance that it travels. Specifically, the time taken, 𝑡, for an object moving at a constant speed, 𝑣, to travel a distance, 𝑑, is equal to 𝑑 divided by 𝑣.

Let’s have a look at a couple more examples.

Example 4: Calculating the Distance Traveled by an Object Moving at a Constant Speed

A train moves uniformly at 12 m/s. The position of the train at two different times is shown. What distance does the train move between the two times?

Answer

This question is asking us to find the distance moved by the train shown in the diagram.

Two different times are shown on the diagram—these are 0 s and 5 s. This means that we know that the train moves for a time of 𝑡=5s.

We are told the speed of the train is 𝑣=12/ms. This means that, in each 1 s interval, the train moves a distance of 12 m. This is illustrated in the diagram below.

The train moves for a total of five 1 s time intervals and moves a distance of 12 m in each of these intervals. Therefore, the total distance moved by the train in the 5 s is equal to 5×12=60.mm

We can get this same result using our formula that relates speed, distance, and time. We can recall that an object that travels a distance 𝑑 in a time 𝑡 has a speed of 𝑣=𝑑𝑡.

In our case, we know the values of 𝑣 and 𝑡, and we are trying to find the value of the distance, 𝑑. This means we need to rearrange the equation to make 𝑑 the subject.

To do this, we first multiply both sides of the equation by 𝑡: 𝑣×𝑡=𝑑×𝑡𝑡.

Then, the 𝑡 in the numerator on the right-hand side cancels with the 𝑡 in the denominator. This gives us 𝑣×𝑡=𝑑, which we can also write as 𝑑=𝑣×𝑡.

We can now substitute in that equation, where 𝑣=12/ms and 𝑡=5s: 𝑑=(12/)×(5).mss

The units on the left- and right-hand sides of the equation need to agree. The speed is measured in metres per second (m/s), and the time is measured in seconds (s), so the distance 𝑑 must have units of metres (m).

Doing the multiplication on the right-hand side of the equation, we find that 𝑑=12×5=60.mm

So, the distance moved by the train is 60 m.

Example 5: Calculating the Time Taken to Travel a Given Distance at a Constant Speed

A train moves a distance of 200 metres at a speed of 25 metres per second. How much time does the train move for?

Answer

This question asks us to find the amount of time for which a train is moving. We are told the speed of the train and the distance that it travels.

We will label the time the train moves for, which is the quantity we are trying to work out, as 𝑡. We will label the speed of the train as 𝑣, which means we know that 𝑣=25/ms. We will label the distance moved by the train as 𝑑, which means we have 𝑑=200m.

We can make a sketch of the situation as follows:

The question asks how much time the train moves for. We could think of this as follows. A speed of 25 m/s means that the train moves a distance of 25 m in each 1 s interval of time.

We could split up the journey of the train into 25 m sections as shown in the diagram below. Only the first few sections are shown.

The question of how much time the train moves for is then the same as asking how many of these 25 m sections there must be in the total 200 m distance.

Mathematically, we can recall that we have a formula relating the speed of an object, 𝑣, the distance it travels, 𝑑, and the time, 𝑡, that it takes to travel this distance: 𝑣=𝑑𝑡.

We know the values of 𝑣 and 𝑑, and we are trying to find the value of 𝑡.

This means we need to rearrange the equation to make 𝑡 the subject.

To do this, we multiply both sides of the equation by 𝑡: 𝑣×𝑡=𝑑×𝑡𝑡.

The 𝑡 in the numerator on the right-hand side cancels with the 𝑡 in the denominator: 𝑣×𝑡=𝑑.

Then, we divide both sides of the equation by 𝑣: 𝑣×𝑡𝑣=𝑑𝑣.

This time, the 𝑣s in the numerator and denominator on the left-hand side cancel out: 𝑡=𝑑𝑣.

We now have an equation for the time, 𝑡, in terms of the distance, 𝑑, and the speed, 𝑣.

Our next task is to substitute in the values given to us in the question. We have 𝑣=25/ms and 𝑑=200m. Substituting these values into our expression for the time 𝑡 gives us: 𝑡=20025/.mms

We have a distance in units of metres and a speed in metres per second. This means we will get a time, 𝑡, with units of seconds.

Doing the division on the right-hand side of the equation, we find that: 𝑡=20025=8.ss

So, the time that the train moves for is 8 s.

It is worth noting that this is equivalent to saying that since the train covers one 25 m section in each 1 s, there must be 8 of the 25 m sections making up the total 200 m distance.

So far, all the distances we have seen have been measured in units of metres and the times have been measured in units of seconds. This has meant that the speeds have units of metres per second.

We initially defined speed as being the distance traveled by an object “per second of time.” A more general definition is that speed is the distance traveled “per unit of time.” We can use whatever units we like for the distance and the time. The units of the speed are then the units we used for the distance divided by the units we used for the time.

The most common choice of units is metres for the distance and seconds for the time, as we have seen so far. However, particularly when talking about longer journeys, it is also common to measure the distance in units of kilometres (km) and the time in units of hours (h).

If we have a distance in kilometres and a time in hours, we can use these to calculate a speed in kilometres per hour (km/h).

Let’s see how this works by considering an example situation.

We will imagine that we have an aircraft that moves at a constant speed, and that it makes a 2‎ ‎000 km journey in a time of 4 h. This is illustrated in the diagram below.

Using this information, we can calculate the speed of the aircraft.

Labeling the speed as 𝑣, the distance as 𝑑, and the time as 𝑡, we can recall that 𝑣=𝑑𝑡.

Substituting in that equation, where 𝑑=2000km and 𝑡=4h, we get 𝑣=20004.kmh

Since the distance has units of kilometres (km), and the time has units of hours (h), we know that the speed must have units of kilometres per hour (km/h).

Evaluating the right-hand side of the expression, we then find that the speed of the aircraft is given by 𝑣=500/.kmh

Let’s now summarize what has been learned in this explainer.

Key Points

  • The speed of an object is a measure of the distance moved by the object in a unit time.
  • An object moving at a constant speed travels equal distances in equal intervals of time.
  • Mathematically, if we label the distance traveled by an object as 𝑑 and the time the object travels for as 𝑡, then the speed, 𝑣, of that object is given by 𝑣=𝑑𝑡.
  • This equation for speed can also be rearranged to make 𝑑 or 𝑡 the subject.
  • Since units on each side of an equation must agree with each other, the units of speed must be units of distance divided by units of time. Common units for speed are metres per second (m/s) or kilometres per hour (km/h).

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