WEBVTT
00:00:01.220 --> 00:00:11.000
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.
00:00:11.470 --> 00:00:16.290
To get started, imagine that you and a friend of yours are both mountaineers.
00:00:16.510 --> 00:00:20.540
You climb some of the world’s tallest peaks for fun and adventure.
00:00:20.980 --> 00:00:31.550
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.
00:00:31.910 --> 00:00:38.160
Just as you’re about to set out on this attempt, a man approaches you and your friend with a wager.
00:00:38.550 --> 00:01:00.520
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.
00:01:01.020 --> 00:01:14.680
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.
00:01:15.330 --> 00:01:22.540
Considering how slow one centimeter per second is, the wager sounds like a pretty good offer to your friend.
00:01:22.960 --> 00:01:24.700
But you’re not so sure about it.
00:01:24.990 --> 00:01:30.280
What can you tell your friend to convince them that this is not a good offer to accept?
00:01:30.820 --> 00:01:36.070
To find out, let’s talk a bit about these two terms, average speed and average velocity.
00:01:36.460 --> 00:01:40.670
We can start out by talking in general about speed and velocity.
00:01:41.020 --> 00:01:44.110
These two terms are different, and they mean different things.
00:01:44.500 --> 00:01:52.310
Speed is a term that depends on distance, while velocity is a term that is based on, or depends on, displacement.
00:01:52.740 --> 00:01:56.340
Distance is a scalar quantity and so is speed.
00:01:56.570 --> 00:01:58.240
Speed cannot be negative.
00:01:58.280 --> 00:02:00.960
It’s always greater than or equal to zero.
00:02:01.480 --> 00:02:06.380
On the other hand, velocity depends not on distance but on displacement.
00:02:07.270 --> 00:02:10.430
Displacement is a vector and so is velocity.
00:02:10.620 --> 00:02:12.590
That is, its sign matters.
00:02:12.590 --> 00:02:14.460
It can be positive or negative.
00:02:15.010 --> 00:02:22.580
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.
00:02:23.130 --> 00:02:28.200
On the map, you’ve drawn in your travel route out as well as your travel route back.
00:02:28.700 --> 00:02:41.520
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.
00:02:41.900 --> 00:02:53.680
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.
00:02:54.220 --> 00:03:01.060
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.
00:03:01.690 --> 00:03:09.350
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.
00:03:09.650 --> 00:03:15.300
But your average velocity assuming you’re able to make it back successfully will be zero.
00:03:15.580 --> 00:03:17.280
You would lose all your gear.
00:03:17.710 --> 00:03:22.700
Let’s go into a little bit more detail about average speed and average velocity.
00:03:23.100 --> 00:03:34.580
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.
00:03:34.970 --> 00:03:40.550
Then we can say that average speed is equal to 𝐷, distance, over Δ𝑡.
00:03:41.080 --> 00:03:46.130
And that average velocity is equal to 𝑑, the displacement, over Δ𝑡.
00:03:46.720 --> 00:03:53.680
Remembering these two relationships will help us keep in mind the differences between average speed and average velocity.
00:03:54.130 --> 00:03:57.620
Let’s get some practice using these equations in a few examples.
00:03:58.780 --> 00:04:05.360
The diagram shows a graph of the variation of the position 𝑥 of an object with time 𝑡.
00:04:05.820 --> 00:04:13.430
What is the velocity of the object in the time interval 𝑡 greater than zero seconds to 𝑡 equal 0.5 seconds?
00:04:13.750 --> 00:04:22.020
What is the velocity of the object in the time interval 𝑡 greater than 0.5 seconds to 𝑡 equals 1.0 seconds?
00:04:22.580 --> 00:04:30.830
What is the velocity of the object in the time interval 𝑡 greater than 1.0 seconds to 𝑡 equals 2.0 seconds?
00:04:31.210 --> 00:04:35.030
In this exercise, we want to solve for three average velocities.
00:04:35.260 --> 00:04:38.300
We’ll call them 𝑣 one, 𝑣 two, and 𝑣 three.
00:04:38.780 --> 00:04:53.690
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.
00:04:54.140 --> 00:05:02.610
To get started, we can recall that average velocity 𝑣 sub avg is equal to displacement divided by time 𝑡.
00:05:02.850 --> 00:05:06.780
Let’s apply this relationship to solve first for 𝑣 sub one.
00:05:07.260 --> 00:05:17.580
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.
00:05:17.960 --> 00:05:24.200
And this difference in position, called the displacement, is then divided by the time it takes to move that distance.
00:05:24.540 --> 00:05:33.900
Looking at our diagram, we can mark out the position of the object at 𝑡 equals 0.5 seconds and 𝑡 equals zero seconds.
00:05:34.300 --> 00:05:44.840
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.
00:05:45.060 --> 00:05:49.130
This means that 𝑣 sub one is 1.0 meters per second.
00:05:49.510 --> 00:05:53.820
That’s the average velocity of the object over this initial time interval.
00:05:54.430 --> 00:05:57.160
Next, we move on to calculating 𝑣 sub two.
00:05:57.490 --> 00:06:05.490
𝑣 sub two is the average velocity of the object between time equals 1.0 seconds and 0.5 seconds.
00:06:05.870 --> 00:06:13.630
Looking at the diagram of the object’s position versus time, we see that its position at both these time values is the same.
00:06:13.680 --> 00:06:15.560
It’s 0.5 meters.
00:06:15.920 --> 00:06:20.250
That means 𝑣 sub two is 0.0 meters per second.
00:06:20.580 --> 00:06:26.150
That’s the average velocity of our object over a time interval where its position doesn’t change.
00:06:26.590 --> 00:06:38.560
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.
00:06:38.880 --> 00:06:50.430
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.
00:06:50.890 --> 00:06:57.110
Calculating this fraction, we find 𝑣 sub three is negative 0.50 meters per second.
00:06:57.500 --> 00:07:01.920
That’s the object’s average velocity on the last leg of its movement.
00:07:02.640 --> 00:07:10.800
We’ve seen in this example how velocity can have both positive and negative values, which is an indication of its vector nature.
00:07:11.380 --> 00:07:17.820
Now, let’s look at an example that compares the vector average velocity with the scalar average speed.
00:07:19.070 --> 00:07:23.860
A helicopter blade spins at exactly 140 revolutions per minute.
00:07:24.280 --> 00:07:28.290
Its tip is 5.00 meters from the center of rotation.
00:07:28.710 --> 00:07:33.090
Calculate the average speed of the blade tip in the helicopter’s frame of reference.
00:07:33.420 --> 00:07:35.460
Give your answer in meters per second.
00:07:36.080 --> 00:07:39.180
What is its average velocity over one revolution?
00:07:39.500 --> 00:07:41.540
Give your answer in meters per second.
00:07:42.220 --> 00:07:48.290
We’re told in this statement that the helicopter blade spins at exactly 140 revolutions per minute.
00:07:48.650 --> 00:07:50.530
We can call that number 𝜔.
00:07:51.040 --> 00:07:58.170
We’re also told that the distance from the axis of the blade’s rotation to its tip is 5.00 meters.
00:07:58.410 --> 00:08:00.670
We’ll call that distance lower case 𝑟.
00:08:00.970 --> 00:08:07.420
In part one, we want to solve for the average speed of the tip of the blade in the helicopter’s frame of reference.
00:08:07.700 --> 00:08:10.270
We’ll call that 𝑠 sub avg.
00:08:10.640 --> 00:08:15.730
And in part two, what we want to solve for its average velocity over one revolution.
00:08:15.920 --> 00:08:18.390
We will call this 𝑣 sub avg.
00:08:18.610 --> 00:08:23.710
To begin on our solution, let’s draw an overhead view of this rotating helicopter blade.
00:08:24.150 --> 00:08:39.810
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.
00:08:40.220 --> 00:08:54.410
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.
00:08:54.830 --> 00:09:04.220
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.
00:09:04.540 --> 00:09:10.170
In our instance, we can say that the distance we’re interested in is one complete revolution.
00:09:10.510 --> 00:09:16.740
That is, the distance that the tip of the blade would travel is two times 𝜋 times the radius of the rotor.
00:09:17.260 --> 00:09:19.740
We’re given that radius in our problem statement.
00:09:20.000 --> 00:09:25.280
So, all this left us to solve for is the time it takes for the blade to make one complete rotation.
00:09:25.680 --> 00:09:35.820
If the rotor makes 140 revolutions every minute, then that tells us that every 60 seconds it rotates 140 times.
00:09:36.040 --> 00:09:43.940
This tells us that Δ𝑡, the time it would take the rotor to go through one complete rotation, is the inverse of this fraction.
00:09:44.290 --> 00:09:53.410
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.
00:09:53.900 --> 00:10:00.220
Now that we’ve solved for Δ𝑡 and we know 𝑟, we’re ready to plug in and solve for our average speed.
00:10:00.740 --> 00:10:09.870
When we enter this expression on our calculator, we find, to three significant figures, that 𝑠 sub avg is 73.3 meters per second.
00:10:10.140 --> 00:10:13.870
That’s the average speed of the tip of the helicopter blade.
00:10:14.310 --> 00:10:22.950
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.
00:10:23.350 --> 00:10:34.410
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.
00:10:34.760 --> 00:10:41.310
With that understanding, we start to see what the average velocity of a rotor blade would be over one revolution.
00:10:41.690 --> 00:10:49.780
When the single rotation begins, the position of our rotor blade is the same as its position when that rotation ends.
00:10:50.020 --> 00:10:58.660
This means that our displacement, which is the difference between those two positions, is equal to zero because those positions are the same.
00:10:59.010 --> 00:11:04.830
This means that our average velocity over one complete rotation is zero meters per second.
00:11:05.060 --> 00:11:12.100
Considered over that time span, the tip of the blade’s displacement is zero and so so is its average velocity.
00:11:14.050 --> 00:11:18.380
Let’s summarize now what we’ve learned about average speed and average velocity.
00:11:18.900 --> 00:11:23.720
Average speed and average velocity are both ways of calculating rates.
00:11:23.930 --> 00:11:25.630
Fundamentally, that’s what they are.
00:11:26.080 --> 00:11:36.710
Second, average speed is based on distance and is a scalar quantity, while average velocity is based on displacement and is a vector quantity.
00:11:37.250 --> 00:11:40.960
This implies that average speed is always nonnegative.
00:11:41.200 --> 00:11:44.330
That is, it’s always greater than or equal to zero.
00:11:44.680 --> 00:11:49.590
While average velocity can be positive or negative or zero as a vector quantity.
00:11:50.290 --> 00:12:00.910
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.