Lesson Explainer: Velocity | Nagwa Lesson Explainer: Velocity | Nagwa

Lesson Explainer: Velocity Mathematics

In this explainer, we will learn how to distinguish between velocity and speed and how to solve problems involving average velocity and average speed.

Speed is a measure of how fast an object moves. The speedometer of a car, for example, tells us how fast the car is going at any given time. That is, the rate at which the car covers a certain distance.

Speed is a scalar quantity and may be measured as a distance per unit of time, such as kilometres per hour (km/h). However, the speed of an object does not tell us the direction in which the object travels. For this we need the velocity. The velocity, which is a vector quantity, gives us both the speed and the direction of travel.

If a car is traveling at 80 km/h, its speed is 80 km/h. However, to say a car travels in a northeasterly direction at 80 km/h gives us a velocity: we have both the speed and the direction of travel of the car.

If we consider an object traveling between two points, there are an infinite number of paths of different lengths that the object could take. The length of a path between two points is the distance an object travels between the points, whereas when considering the velocity of an object, we refer to its displacement. The displacement measures how far, and in what direction, the second point is from the first.

Suppose a bird flies 30 km due east and then 40 km due north.

The total distance, 𝐷, covered by the bird is the sum of the distances in each of the two stages of its journey. That is, 𝐷=𝐴𝐵+𝐵𝐶=30+40=70km. On the other hand, the bird’s displacement is represented by the directed line segment 𝐴𝐶. By the Pythagorean theorem, its magnitude is 𝐴𝐶=𝐴𝐵+𝐵𝐶=30+40=2500=50.km

Hence, while the distance covered by the bird’s flight is 70 km, the magnitude of the bird’s displacement, that is, how far it is from where it started, is 50 km. We know also that tan𝜃=4030, and taking the inverse tangent we find 𝜃=53.130. Converting to degrees, minutes, and seconds, this is 53748. The bird’s displacement is, therefore, 50 km, 53748 in the northeasterly direction: it has both a magnitude and a direction.

We use distance to define speed and displacement to define velocity, as follows.

Definitions: Speed and Velocity

Speed, which is a scalar quantity, is the rate at which an object covers distance: speeddistancetime=.

Velocity, which is a vector quantity, is the magnitude and direction of an object’s change in position: velocitydisplacementtime=.

Now, if we consider again motion in multiple stages, for example, the journey of our bird above, we may think in terms of the average speed. We define this as follows.

Definition: Average Speed

The average speed an object travels is equal to the total distance covered, over the total time taken: averagespeedtotaldistancetotaltime=.

This is a scalar quantity measured as a distance per unit of time, for example, kilometres per hour (km/h).

Returning to our bird example above, suppose the first stage (30 km) takes the bird three quarters of an hour and the second stage (40 km) takes the bird an hour and a quarter. We then have averagespeedtotaldistancetotaltimekmh==30+400.75+1.25=702=35/.

Now, recall that the velocity of an object is a vector quantity, measuring both the displacement and the direction of the object’s motion. The average velocity, also a vector quantity, is defined as follows.

Definition: Average Velocity

The average velocity of an object’s motion is equal to the net displacement, divided by the total time: averagevelocitynetdisplacementtotaltime=, where the net displacement is the displacement measured directly from the object’s starting position to its end position. Since displacement is a vector quantity, average velocity is also a vector quantity and so may be positive or negative. Its magnitude, which is a scalar, is measured as a distance per unit of time and is therefore always positive.

Going back to our bird, we recall that the total displacement was 50 km. Hence, averagevelocitynetdisplacementtotaltimekmhinthenortheasterlydirection==502=25/,.

We note that this is less than the bird’s average speed, which was 35 km/h. This is because in the numerator for average speed, the total distance is calculated by summing the distances of the two stages of the journey, whereas in the numerator for average velocity, the net displacement is the direct displacement between the start and finish positions.

Let’s look at some examples.

Example 1: Calculating Average Velocity of a Multipart Journey Given Distances and Speeds

Calculate the average velocity of a car which moved in a straight line a distance of 120 m with a velocity of 8 m/s and then moved in the same direction on the same line a distance of 180 m with a velocity of 6 m/s. Give your answer in m/s to 1 decimal place.

Answer

We are given the distances, a direction, and the speeds for the two stages of the car’s journey.

To calculate the average velocity of the car, we will use the formula averagevelocitynetdisplacementtotaltime=.

In this case, since there was no change in direction, we note that the net displacement was equal to the total distance traveled, in the direction of the straight line. Letting 𝑑 and 𝑑 be the distances traveled in stages 1 and 2, then the magnitude of the net displacement of the car was 𝑑+𝑑=120+180=300m.

To work out the total time, we will need to calculate the times taken, 𝑡 and 𝑡, for each stage of the journey. To do this we use the fact that time is equal to distance over speed: 𝑡=𝑑𝑠. In stage 1, the speed 𝑠=8/ms, and in stage 2, the speed 𝑠=6/ms. Hence, timetakenforstagestimetakenforstages1=𝑡=𝑑𝑠=1208=15,2=𝑡=𝑑𝑠=1806=30.

The total time taken by the car was, therefore, 15+30=45s.

Hence, the average velocity of the car was as follows: averagevelocitymstod.p.,inthegivendirection=30045=6.7/1.

Example 2: Calculating Average Velocity of a Multipart Journey Given Times and Speeds

An object moves north at 12 m/s for 10 seconds and then stops and stays motionless for 10 seconds before moving north at 12 m/s for another 10 seconds. What is the object’s average velocity?

Answer

In this example, there are three stages to the motion of the object.

In the first and third stages, the object moves in the positive, that is, the northward, direction, and in the second stage, the object is at rest. To find the object’s average velocity we will use the formula averagevelocitynetdisplacementtotaltime=.

In this case, since there is no change in direction, we note that the magnitude of the net displacement is equal to the total distance traveled. Our first step then is to find the distances traveled in each stage of the journey. To do this, we use the fact that distance = speed × time: distancestagemdistancestagemdistancestagem1=12×10=120,2=0×10=0,3=12×10=120.

The total distance covered in the journey is, therefore, 120+0+120=240m. Since each of the three stages takes 10 seconds, the total time is 10+10+10=30s. The object’s average velocity, which is in a northward direction, is then given by averagevelocityms=24030=8/.

We might also have considered this problem using a displacement–time graph, where each line segment represents a particular stage of the object’s motion, as shown below.

With time and displacement on the horizontal and vertical axes, respectively, we see, for example, that, starting from time 𝑡=0seconds, after 10 seconds the object has moved 120 m and has the coordinates (10,120). The line segments representing positive motion have a positive slope and the middle segment representing rest has zero slope.

Now, recall that the slope, 𝑚, of the line between two points (𝑥,𝑦) and (𝑥,𝑦) is given by 𝑚=𝑦𝑦𝑥𝑥.

If we use the start and finish positions of our object as the two points on the line between them, we can find the object’s average velocity by calculating the slope of this line.

In our case, our points have the coordinates (0,0) and (30,240). Hence, the slope of the line, which is the average velocity between the start and finish positions, is given by 𝑚==2400300=24030=8/.averagevelocityms

Let’s look again at the second method we used to calculate the average velocity in the previous example.

With time on the horizontal axis and displacement on the vertical axis, we plotted the uniform motion, that is, the motion of an object in fixed directions, with constant speeds. Each line segment of the graph represented a stage of the motion, illustrating the displacement, if any, in a particular direction, in a specified amount of time.

This type of graph is called a displacementtime graph. Consider the example below.

From point 𝐴 to point 𝐵, an object moves a distance of 20 m, in 4 seconds. Since the slope of the line segment is positive, we know that the motion is in the positive (or forward) direction. From point 𝐵 to point 𝐶, the object then travels, again forward, 30 m in 2 seconds. From point 𝐶 to point 𝐷 it is at rest for 2 seconds, then from point 𝐷 to point 𝐸, since the gradient is negative, we know it goes back 20 m in 4 seconds.

Now, recall that speeddistancetime=. So, for example, from point 𝐴 to point 𝐵, the object’s speed was 204=5/ms. Recall also that the distance traveled from one point to another with a direction, the displacement, is to velocity what distance is to speed: velocitydisplacementtime=. So, for example, the velocity while going from 𝐷 to 𝐸 is velocityms=204=5/, whereas the speed while going from 𝐷 to 𝐸 is speedms=204=5/, and we recognize that the speed is the magnitude of the velocity.

Let’s now consider the average velocity, that is, the net displacement, which is the displacement between the overall start and finish positions, divided by the total time taken. In this example, the net displacement in going from 𝐴 to 𝐸 is 30 m in the positive direction. The time taken for this to occur is 12 seconds.

In terms of the line between 𝐴 and 𝐸, in the Cartesian plane, this is the change in 𝑦 over the change in 𝑥, which is precisely the slope, or gradient, of the line between points 𝐴 and 𝐸. 𝑚==300120=3012=2.5/.averagevelocitymsinthepositivedirection

How To: Using Displacement–Time Graphs to Calculate Velocity

For an object whose motion begins at point 𝐴(𝑡,𝑑) and ends at point 𝐵(𝑡,𝑑), on a displacement–time graph, the object’s average velocity is the slope of the line from point 𝐴 to point 𝐵, that is, the overall change in position, or the net displacement, over the total time taken: averagevelocitynetdisplacementtotaltime==𝑑𝑑𝑡𝑡.

In our next example, we consider displacements and average velocity for a multistage journey involving a change in direction.

Example 3: Exploring Displacements, Times, and Velocities of a Multipart Journey

A fish swims a distance of 15 m to the left at an average speed of 2 m/s and then immediately turns around and swims half this distance to the right. The time that passes while all this occurs is 12 s, as shown in the diagram. In this question, consider displacement to the left as positive.

  1. What is the displacement of the fish from its starting point 12 s after it starts moving?
  2. For how long does the fish swim to the left?
  3. What is the fish’s average velocity while it moves to the right? Answer to one decimal place.
  4. What is the average velocity of the fish over the 12 s that the fish moves for?

Answer

Part 1

We are asked first for the displacement of the fish after 12 seconds and we are told that 12 s is the total time taken for the fish’s entire journey. Hence, we need to find the net displacement; that is, how far, and in what direction, the fish is from the starting point when it completes its journey.

There are two stages to the journey. The first is 𝑑=15m left (which is specified as the positive direction), followed by 𝑑2=152=7.5m right, that is, in the negative direction. The net displacement is then netdisplacementminthepositivedirection=157.5=7.5.

Part 2

We are next asked to find how long the fish swims to the left. We are told that the distance covered in this leg of the journey is 15 m, at an average speed of 2 m/s. To find the time taken, we therefore use the formula timedistancespeed=.

In our case, this is times=152=7.5.

Part 3

Next, we are asked to find the fish’s average velocity while it moves to the right, that is, in the negative direction. To do this, we will use the formula averagevelocitynetdisplacementtotaltime=.

The fish’s displacement to the right is 𝑑2=152=7.5m, where the negative sign indicated the direction.

Now, to find the denominator in our formula, that is, the time taken for this leg of the fish’s journey, we note that if the first stage took 7.5 s, as we found above, and the total time taken for the whole journey was 12 s, then the time taken for the second stage is 127.5=4.5s. We now have everything we need to calculate the average velocity as the fish moves to the right: averagevelocitymstod.p.=7.54.5=1.7/1

Part 4

Finally, we are asked to find the average velocity of the fish over the 12 s that it moves for. Again, we will use the formula averagevelocitynetdisplacementtotaltime=, where, now, the net displacement is the displacement from its starting position to its final position. We found this in the first part of this question to be 7.5 m. The total time taken overall was 12 s. Hence, averagevelocityms=7.512=0.625/.

We could also consider this problem using a displacement–time graph, where each line segment represents a particular stage of the fish’s motion, as shown below.

With time on the horizontal and displacement on the vertical axis, we see, for example, that, starting from time 𝑡=0seconds, after 7.5 seconds, the fish has moved 15 m. The line segment representing this stage has a positive slope indicating motion in a forward, or positive, direction.

Now, recall that the slope, 𝑚, of the line between two points (𝑥,𝑦) and (𝑥,𝑦) is given by 𝑚=𝑦𝑦𝑥𝑥.

If we use the start and finish positions of our fish as the two points on the line between them, we can find the fish’s average velocity by calculating the slope of this line.

In our case, our points have the coordinates (0,0) and (12,7.5). Hence, the slope of the line, which is the average velocity between the fish’s start and finish positions, is given by 𝑚==7.50120=7.512=0.625/.averagevelocityms

The next example also considers average velocity for a multistage journey.

Example 4: Exploring Distances, Times, and Velocities of a Multipart Journey

A person is late for an appointment at an office that is at the other end of a long straight road to his home. He leaves his house and runs toward his destination for a time of 45 seconds before realizing that he has to return home to pick up some documents that he will need for the appointment. He runs back home at the same speed he ran at before and spends 185 seconds looking for the documents and then runs toward his appointment again. This time, he runs at 5.5 m/s for 260 seconds and then arrives at the office.

  1. How much time passes between the person first leaving his house and arriving at his appointment?
  2. What is the distance between the person’s house and his office?
  3. What is the person’s average velocity between first leaving his house and finally arriving at the office? Give your answer to two decimal places.

Answer

The person’s journey consists of four stages: the first two are of the same distance for 45 s, but in opposite directions; the third covers no distance for 185 s; and the final leg lasts 260 s at a speed of 5.5 m/s.

Part 1

We’re asked first to find how much time it takes from setting off to arriving at his appointment. For this, we sum the times of each stage of the journey: timetakens=45+45+185+260=535.

Part 2

Next, we are asked to find the distance between the person’s house and his office. To find this, we may simply use the information we have on the last leg of his journey, since he covers the whole distance in this stage. We use the formula distancespeedtime=×.

Since we know that the speed for this leg is 5.5 m/s and the time taken is 260 s, the distance is given by distancem=5.5×260=1430.

Part 3

Finally, we are asked for the person’s average velocity between first leaving the house and finally arriving at the office. For this, we use the formula averagevelocitynetdisplacementtotaltime=.

The net displacement is the displacement whose magnitude is the direct distance between his home and the office. We have just found this to be 1‎ ‎430 m, and we know that the total time taken overall was 535 s. Hence, averagevelocitymstod.p.=1430535=2.67/,2

That is, if the positive direction is from home toward the office, to two decimal places, the person’s average velocity between first leaving his house and finally arriving at the office is 2.67 m/s.

Our next example demonstrates how to find both the average speed and the average velocity for a moving object.

Example 5: Calculating Distance, Displacement, Speed, and Velocity for a Moving Object

A child kicks a ball and it moves along horizontal ground toward a wall 2 m away. The ball moves at 2.25 m/s toward the wall, hits the wall, and bounces half the distance straight back to the child but moving only at an average speed of 1.75 m/s after hitting the wall.

  1. What is the total distance that the ball moved?
  2. What was the ball’s net displacement toward the wall?
  3. How much time did the ball move for? Give your answer to two decimal places.
  4. What was the ball’s average speed during its motion to the nearest metre per second?
  5. What was the ball’s average velocity toward the wall from when the ball was kicked to when it had rolled back from the wall and come to a stop? Give your answer to two decimal places.

Answer

There are two stages in the motion of the ball. The first is a distance of 2 m over horizontal ground at 2.25 m/s, and the second covers half the original distance in the opposite direction at an average speed of 1.75 m/s.

Part 1

The total distance the ball moved is the sum of the forward (2 m), and backward (12×2=1m) distances; that is, 2+1=3m.

Part 2

The ball’s net displacement is the displacement between where the ball started and where it ended up. That is, netdisplacementmforwardmbackmforward=21=1.

Part 3

To find the total time for which the ball was in motion, we will need to find the times of each stage of its journey. To do this, we use the formula timedistancespeed=.

Letting 𝑡 and 𝑡 be the times for each stage, 𝑑 and 𝑑 be the distances for each stage, and 𝑠 and 𝑠 be the speeds for each of the two stages, then we have 𝑡=𝑑𝑠=22.25=0.8888,𝑡=𝑑𝑠=11.75=0.5714.

The total time the ball was in motion is, therefore, 𝑡+𝑡=0.888+0.5714=1.46seconds, to 2 d.p.

Part 4

We are next asked to find the ball’s average speed during its motion to the nearest metre per second. The average speed is given by averagespeedtotaldistancetotaltime=, which in our case is averagespeed=31.46=2.055.

To the nearest metre per second, the average speed is, therefore, 2 m/s.

Part 5

Finally, the ball’s average velocity from when it was kicked to its final position is given by averagevelocitynetdisplacementtotaltime=.

We have already found the net displacement to be 1 m toward the wall. Hence, averagevelocitymstod.p.=11.46=0.68/2, in the direction from the boy toward the wall.

We could also calculate the average velocity for this problem using a displacement–time graph, where each line segment represents a particular stage of the ball’s motion, as shown below.

With time on the horizontal and displacement on the vertical axis, we see, for example, that, starting from time 𝑡=0seconds, after approximately 0.89 seconds, the ball moved 2 m. The line segment representing this stage has a positive slope indicating motion in a forward, or positive, direction.

Now, recall that the slope, 𝑚, of the line between two points (𝑥,𝑦) and (𝑥,𝑦) is given by 𝑚=𝑦𝑦𝑥𝑥.

If we use the start and finish positions of our ball as the two points on the line, we can find the ball’s average velocity by calculating the slope of this line.

In our case, our points have the coordinates (0,0) and (1.46,1). Hence, the slope of the line, which is the average velocity between the start and finish positions, is given by 𝑚==101.460=11.46=0.68/2averagevelocitymstod.p.

In our final example, we find the average velocity where there is a change in the direction of motion. This is a similar scenario to the example with the bird flying east and then north, discussed at the beginning in this explainer.

Example 6: Calculating Average Velocity of a Multipart Journey with a Change in Direction

A man walked 6 km in an easterly direction for 1.2 hours. He then walked 8 km in a northerly direction for 2 hours. Calculate the magnitude of the average velocity of the man.

Answer

The man’s journey consists of two stages.

We are given the distance covered in each stage, that is, 6 km and 8 km, and the time the man took for each stage, 1.2 hours and 2 hours, respectively. To find the magnitude of the average velocity of the man, we will use the formula averagevelocitynetdisplacementtotaltime=.

For the denominator, we simply sum the times for the two stages. Hence, the total time is 1.2+2=3.2hours. For the numerator, we will need to find the net displacement, that is, the displacement between the initial and final positions.

We can use the Pythagorean theorem to find this as follows: netdisplacementkm=6+8=100=10.

Technically, since displacement is a vector quantity, we should specify a direction of displacement. To do this, we could calculate the angle between the horizontal and the direction of displacement, or, alternatively, we might note that the displacement is in an approximately northeasterly direction. However, since we are asked to find the magnitude of the average velocity, this is not strictly necessary here.

Now, using our formula, the magnitude of the average velocity is ||=103.2=3.1/.averagevelocitykmh

It is worth reminding ourselves that we have been discussing situations throughout where for each stage of an object’s motion, the velocity was constant. As we have seen on a displacement–time graph, a constant velocity is the slope of the line segment representing that stage of the motion.

Now, suppose instead that the velocity is not constant. The graph of the object’s motion is then a curve. In the example shown below, the motion has nonconstant velocity in the positive direction.

For a point on the curve, we may find what is called the instantaneous velocity by calculating the slope of the tangent at that point.

Although beyond the scope of this explainer, we may still also find the average velocity for the whole or part of an object’s motion. We do this by finding the slope of the line, either from the start and finish positions or between two points during the motion.

Let’s recap some of the key points on average speed and average velocity covered in this explainer.

Key Points

  • Speed is a scalar quantity and is the rate at which an object covers distance, where distance is the length of a path between two points: speeddistancetime=. Speed is measured as a distance per unit of time.
  • Velocity, which is a vector quantity, specifies both the magnitude and the direction of an object’s change in position. velocitydisplacementtime=. Given two points, displacement measures how far, and in which direction, the second point is from the first. Velocity is specified as a distance per unit of time and has a direction.
  • The average speed of an object’s motion is a scalar quantity also measured as a distance per unit of time: averagespeedtotaldistancetotaltime=.
  • The average velocity of an object’s motion is given by averagevelocitynetdisplacementtotaltime=, where net displacement is the object’s displacement from its starting point to its finishing point. This is a vector quantity, so it has a direction and is measured as a distance per unit of time.
  • On a displacement–time graph, the average velocity of an object’s motion is the slope, or gradient, of the line between the object’s start and finish positions.

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