In this video, we’re going to learn about average speed and velocity, what these two terms mean, how they’re different from one another, and how to use them practically.
To get started, imagine that you and a friend of yours are both mountaineers. You climb some of the world’s tallest peaks for fun and adventure. You’re both in the final stages of preparation for your toughest climb yet, an ascent of Mount Giant, one of the world’s tallest and most secret peaks. Just as you’re about to set out on this attempt, a man approaches you and your friend with a wager.
The man, who is the owner of the house where the two of you have been staying, says that if both of you are able to successfully climb Mount Giant and return and have an average velocity of greater than or equal to one centimeter per second, then both of you will be allowed to stay at his lodge for free for the rest of your life. But, he says, if your average velocity is less than one centimeter per second on the trip, then you must give him all of your mountaineering equipment, your ropes, your harnesses, your crampons, everything.
Considering how slow one centimeter per second is, the wager sounds like a pretty good offer to your friend. But you’re not so sure about it. What can you tell your friend to convince them that this is not a good offer to accept? To find out, let’s talk a bit about these two terms, average speed and average velocity. We can start out by talking in general about speed and velocity. These two terms are different, and they mean different things.
Speed is a term that depends on distance, while velocity is a term that is based on, or depends on, displacement. Distance is a scalar quantity and so is speed. Speed cannot be negative. It’s always greater than or equal to zero. On the other hand, velocity depends not on distance but on displacement. Displacement is a vector and so is velocity. That is, its sign matters. It can be positive or negative.
Imagine that you and your friend had a topographic map that show the route that both of you travelled on a recent ascent up a mountain. On the map, you’ve drawn in your travel route out as well as your travel route back. The average speed at which you moved during this expedition is equal to the total distance you travelled, we can call it capital 𝐷, divided by the time it took you to get up and back down a mountain.
On the other hand, because you started and ended in the same location, your displacement for the journey is zero and that means so is your average velocity for the trip. As you think back to the wager offered to you and your friend, now you understand better why you didn’t want to accept it. If the man was talking about average speed, you would be confident that you could meet that standard of one centimeter per second or more. But your average velocity assuming you’re able to make it back successfully will be zero. You would lose all your gear.
Let’s go into a little bit more detail about average speed and average velocity. If we let capital 𝐷 equal distance travelled, lowercase 𝑑 be displacement, a vector, and Δ𝑡 be the time elapsed in order to make our journey. Then we can say that average speed is equal to 𝐷, distance, over Δ𝑡. And that average velocity is equal to 𝑑, the displacement, over Δ𝑡. Remembering these two relationships will help us keep in mind the differences between average speed and average velocity. Let’s get some practice using these equations in a few examples.
The diagram shows a graph of the variation of the position 𝑥 of an object with time 𝑡. What is the velocity of the object in the time interval 𝑡 greater than zero seconds to 𝑡 equal 0.5 seconds? What is the velocity of the object in the time interval 𝑡 greater than 0.5 seconds to 𝑡 equals 1.0 seconds? What is the velocity of the object in the time interval 𝑡 greater than 1.0 seconds to 𝑡 equals 2.0 seconds?
In this exercise, we want to solve for three average velocities. We’ll call them 𝑣 one, 𝑣 two, and 𝑣 three. These average velocities are based on our position versus time graph, where 𝑣 one applies to the first leg of the journey, 𝑣 two applies to the second leg, and 𝑣 three applies to the third and final leg of this movement. To get started, we can recall that average velocity 𝑣 sub avg is equal to displacement divided by time 𝑡. Let’s apply this relationship to solve first for 𝑣 sub one.
𝑣 sub one is equal to the position of our object 𝑝 when time equals 0.5 seconds minus our object’s position when time equals zero seconds. And this difference in position, called the displacement, is then divided by the time it takes to move that distance. Looking at our diagram, we can mark out the position of the object at 𝑡 equals 0.5 seconds and 𝑡 equals zero seconds. And if we drop a vertical line down from our second point, we see that the time interval Δ𝑡 for all this to happen is equal to 0.5 seconds. This means that 𝑣 sub one is 1.0 meters per second. That’s the average velocity of the object over this initial time interval.
Next, we move on to calculating 𝑣 sub two. 𝑣 sub two is the average velocity of the object between time equals 1.0 seconds and 0.5 seconds. Looking at the diagram of the object’s position versus time, we see that its position at both these time values is the same. It’s 0.5 meters. That means 𝑣 sub two is 0.0 meters per second. That’s the average velocity of our object over a time interval where its position doesn’t change.
Finally, we move on to calculating 𝑣 three, the average velocity of our object over the third time interval from time 𝑡 equals 2.0 seconds to 1.0 seconds. Looking on our diagram, we see that at 2.0 seconds our object has a position of 0.0 meters and at 1.0 seconds it has a position of 0.5 meters. Calculating this fraction, we find 𝑣 sub three is negative 0.50 meters per second. That’s the object’s average velocity on the last leg of its movement.
We’ve seen in this example how velocity can have both positive and negative values, which is an indication of its vector nature. Now, let’s look at an example that compares the vector average velocity with the scalar average speed.
A helicopter blade spins at exactly 140 revolutions per minute. Its tip is 5.00 meters from the center of rotation. Calculate the average speed of the blade tip in the helicopter’s frame of reference. Give your answer in meters per second. What is its average velocity over one revolution? Give your answer in meters per second.
We’re told in this statement that the helicopter blade spins at exactly 140 revolutions per minute. We can call that number 𝜔. We’re also told that the distance from the axis of the blade’s rotation to its tip is 5.00 meters. We’ll call that distance lower case 𝑟. In part one, we want to solve for the average speed of the tip of the blade in the helicopter’s frame of reference. We’ll call that 𝑠 sub avg. And in part two, what we want to solve for its average velocity over one revolution. We will call this 𝑣 sub avg.
To begin on our solution, let’s draw an overhead view of this rotating helicopter blade. Looking down from an aerial view on this rotating helicopter rotor, we see that it’s rotating with an angular speed 𝜔 of 140 rpm and that each blade on the rotor has a length 𝑟 of 5.00 meters. If we mark the tip of one of the blades of the rotor and follow its path as it moves around a revolution, we want to solve for the average speed of that tip of the blade as well as its average velocity.
Starting with average speed, we can recall that the average speed of an object is equal to the distance it travels divided by the time it takes to travel that distance. In our instance, we can say that the distance we’re interested in is one complete revolution. That is, the distance that the tip of the blade would travel is two times 𝜋 times the radius of the rotor. We’re given that radius in our problem statement. So, all this left us to solve for is the time it takes for the blade to make one complete rotation.
If the rotor makes 140 revolutions every minute, then that tells us that every 60 seconds it rotates 140 times. This tells us that Δ𝑡, the time it would take the rotor to go through one complete rotation, is the inverse of this fraction. In other words, in sixty one hundred and fortieths of a second, the rotor goes through a complete rotation, or every three-sevenths of a second.
Now that we’ve solved for Δ𝑡 and we know 𝑟, we’re ready to plug in and solve for our average speed. When we enter this expression on our calculator, we find, to three significant figures, that 𝑠 sub avg is 73.3 meters per second. That’s the average speed of the tip of the helicopter blade.
Now, we move on to solving for the average velocity of the tip of the blade over one rotation, which we’ll find is not the same as average speed. Recall that average velocity is equal to displacement over time, where displacement is the difference in position from the start to the end of an object’s motion. With that understanding, we start to see what the average velocity of a rotor blade would be over one revolution.
When the single rotation begins, the position of our rotor blade is the same as its position when that rotation ends. This means that our displacement, which is the difference between those two positions, is equal to zero because those positions are the same. This means that our average velocity over one complete rotation is zero meters per second. Considered over that time span, the tip of the blade’s displacement is zero and so so is its average velocity.
Let’s summarize now what we’ve learned about average speed and average velocity. Average speed and average velocity are both ways of calculating rates. Fundamentally, that’s what they are. Second, average speed is based on distance and is a scalar quantity, while average velocity is based on displacement and is a vector quantity.
This implies that average speed is always nonnegative. That is, it’s always greater than or equal to zero. While average velocity can be positive or negative or zero as a vector quantity. Keep in mind that average speed is a scalar based on distance, while average velocity is a vector based on displacement, and the difference between the two will always be clear.