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In this video, we’re going to learn about velocity vectors: What they are?
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What information they show us?
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And how we can combine them properly?
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Imagine that you are an airplane pilot and your work takes you to various islands throughout the Pacific Ocean.
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After learning about a shipwreck on one particular island, you fly a mission to drop vital food and supplies to the survivors on the island.
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As you head due east, the island is to the northeast of your current bearing.
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As you fly along, a steady wind out of the north begins to affect your flight path.
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You want to figure out, given this wind, what your velocity vector should be so that you head straight towards the island and you are able to get there as soon as possible.
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To figure out this direction, we’ll need to understand velocity vectors.
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When we talk about velocity vectors, we’re talking about something that shows an object’s instantaneous speed as well as its direction.
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Imagine we had a three-dimensional coordinate space, where our dimensions were velocity in the 𝑥-direction, velocity in the 𝑦-direction, and velocity in the 𝑧-direction.
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If we picked a particular point in that space, it would have components: 𝑣 sub 𝑥, 𝑣 sub 𝑦, and 𝑣 sub 𝑧.
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If we wanted to write that point as a velocity vector, we would write that it is equal to 𝑣 sub 𝑥 in the 𝑖 direction, 𝑣 sub 𝑦 in the 𝑗, and 𝑣 sub 𝑧 in the 𝑘.
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We’ve said that a velocity vector shows two things: an object’s speed and its direction.
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In terms of direction, that’s given by the unit vectors 𝑖, 𝑗, and 𝑘 in this expression.
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The object’s speed is given by the magnitude of the velocity vector — that is, its length.
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So if we have the components of the velocity 𝑣, we can insert those, square them, add them together, and take their square root to find speed.
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That’s a bit about individual velocity vectors.
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Now, let’s consider what will happen if we add them together.
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In general, it’s the case that vectors can be added together if they have the same units.
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For example, if a walker and a runner were approaching one another at different velocities and we called the walker’s velocity of one meter per second to the right 𝑣 sub one and the runner’s velocity of two meters per second to the left 𝑣 sub two.
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Then we could solve for a total velocity by adding 𝑣 sub one and 𝑣 sub two.
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Because these two vectors have the same units, meters per second, we can combine them and find, if we want to graphically, the resultant total velocity vector.
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Let’s get some practice adding velocity vectors together through an example.
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An athlete crosses a 30.0-meter-wide river by swimming perpendicular to the water current at a speed of 0.500 meters per second relative to the water.
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He reaches the opposite side at a distance 75.0 meters downstream from his starting point.
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How fast is the water in the river flowing with respect to the ground?
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What is the speed of the swimmer with respect to a friend at rest on the bank of the river?
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We want to solve in this exercise for two things: first, the speed of the water in the river with respect to the ground.
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We’ll call that 𝑣 sub 𝑤.
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Second, we want to solve for the speed of the swimmer with respect to a stationary friend on the bank.
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We’ll call that speed 𝑣 sub 𝑠.
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We’re told the river is 3.0 [30.0] meters wide and that the swimmer swims at 0.500 meters per second relative to the water and also that he ends up at distance 75.0 meters downstream from his starting point.
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Let’s draw a diagram of this situation to start our solution.
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We have a river that’s 30.0 meters wide.
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We can call that width 𝑤.
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Our swimmer sets out from the left bank swimming at a speed we’ve called 𝑣 with respect to the water, where 𝑣 is 0.500 meters per second.
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If the river had no current, the swimmer would end up on the opposite bank directly across from where he started.
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But there is a current in the river moving with a speed we’ve called 𝑣 sub 𝑤.
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This affects the trajectory of the swimmer so that the actual path he follows is a diagonal line.
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And he reaches the far shore at distance we’ve called 𝑑 below the horizontal line of where he started.
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That distance is 75.0 meters.
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Knowing all this, we want to solve for the current speed of the river.
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If we were to zoom in on the swimmer, we would see that there are two velocity vectors affecting the swimmer’s motion: one we’ve called 𝑣, which is the swimmer’s self-powered motion across the river; the other is the river current pushing the swimmer downstream.
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To solve for the speed of the current, we want to remember an important fact about motion in multiple directions.
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And that is that motion in perpendicular directions is independent.
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In our case, this means we can figure out how long it takes the swimmer to cross the river just by looking at the swimmer’s own velocity 𝑣.
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That time to cross the river is unaffected by 𝑣 sub 𝑤 because 𝑣 sub 𝑤 is acting perpendicular to 𝑣.
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If we recall that for a constant speed, the speed is equal to the distance travelled divided by the time it took to travel that distance, we can write that 𝑣 — the swimmer’s speed — is equal to 𝑤 — the width of the river — divided by 𝑡 — the time it takes for the swimmer to cross the river.
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That means that the time to cross is equal to 𝑤 divided by 𝑣 or 30.0 meters divided by 0.500 meters per second, which is equal to 60.0 seconds.
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That’s how long it takes the swimmer to reach the other side.
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We can reuse this expression 𝑣 equals 𝑑 over 𝑡 now to solve for 𝑣 sub 𝑤.
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𝑣 sub 𝑤, the speed of the current, is equal to the distance downstream the swimmer ended up on the opposite side divided by the total time it took to cross the river.
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75.0 meters divided by 60.0 seconds is equal to 1.25 meters per second.
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That’s how fast the current in the river is flowing.
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Then, we imagine that a friend of the swimmer is standing still on shore and watching as the river is crossed.
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We want to solve for the speed of the swimmer relative to that stationary friend.
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If we look back at a diagram of the velocity vectors acting on the swimmer, we see that we can combine them to form a net velocity vector, which we’ve called 𝑣 sub 𝑠.
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It’s the addition of 𝑣 and 𝑣 sub 𝑤 as vectors.
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𝑣 sub 𝑠 is a speed.
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That is, it’s the magnitude of the addition of 𝑣 and 𝑣 sub 𝑤.
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So to find 𝑣 sub 𝑠, we’ll take 𝑣 sub 𝑤, square it, add it to 𝑣 squared, and then take the square root of that sum.
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When we plug in the given value for 𝑣 and the value for 𝑣 sub 𝑤 we solved for and enter this expression on our calculator, we find that 𝑣 sub 𝑠, to three significant figures, is 1.35 meters per second.
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That’s the speed of the swimmer under the influence of the river current relative to a stationary observer on shore.
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In our opening example, we talked about how an airplane pilot will need to pick a bearing based on current weather conditions.
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Let’s do an example now that involves solving for that sort of information.
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A small plane can fly at 175 kilometers per hour in still air.
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The plane flies in a wind that blows directly out of the west at 36 kilometers per hour.
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At what angle west of north must the plane point in order for it to move directly north?
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How much time is needed for the plane to reach a point 300 kilometers directly north of its current position?
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Let’s call the angle representing the bearing of the plane 𝜃.
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We also want to solve for the amount of time needed for the plane to travel 300 kilometers directly north from where it currently is.
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We’ll call this time 𝑡.
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In this statement, we’re given the plane’s speed, 175 kilometers per hour in still air.
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And we’re given the velocity vector of the wind.
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We’re told it blows out of the west at a speed of 36 kilometers per hour.
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Let’s start on our solution by drawing a diagram of the plane in travel and this wind affecting its motion.
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If we draw in our four compass directions: north, south, east, and west, we’re told there’s a wind coming out of the west.
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We’ve called the speed of that wind 𝑣 sub 𝑤, given as 36 kilometers per hour.
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We’re told that our plane wants to be able to fly in a direction due north including the effect of the wind on the plane’s flight.
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To do that, the plane will need to pick its velocity vector.
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That is, fly in such a direction that the western component of its velocity vector is counteracted by the eastern pushing of the wind.
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When this happens, the plane’s net motion will be to the north — that is the vector sum of these two velocities.
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We’re given the magnitude of the plane’s velocity, its speed which we’ve called 𝑣 sub 𝑝 given as 175 kilometers per hour.
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To solve for 𝜃, we want to consider the component of the plane’s velocity that will counteract 𝑣 sub 𝑤.
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We can write this as an equation.
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We write that 𝑣 sub 𝑝, the plane’s speed, multiplied by the sine of the angle 𝜃 is equal to 𝑣 sub 𝑤.
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This is the condition we enforce so that the plane’s net motion can be directly north.
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Dividing both sides by 𝑣 sub 𝑝 and then taking the arcsine of both sides of the equation, we find that 𝜃 is equal to the inverse sine of the speed of the wind divided by the speed of the plane.
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When we plug in for those values — 36 kilometers per hour for the wind speed and 175 kilometers per hour for plane speed — and calculate this expression, we find, to two significant figures, that 𝜃 is 12 degrees.
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That’s the direction west of north at which the plane should head so that its net motion is directly north.
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Heading this way, we then imagine that the plane flies a journey of 300 kilometers north.
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And we want to solve for the time that this journey would take.
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To solve for this time, we can recall that average speed 𝑣 is equal to the distance travelled divided by the time it takes to travel that distance.
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We can rearrange this expression so that it reads 𝑡 is equal to 𝑑 over 𝑣.
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In our case, 𝑑 is 300 kilometers and 𝑣 is 𝑣 sub 𝑝, the plane’s speed, multiplied by the cosine of 𝜃.
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That’s the northerly component of the plane’s velocity.
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Plugging in our values for 𝑣 sub 𝑝 and 𝜃, when we calculate this fraction, we find it’s 1.75 hours or to, one significant figure, 100 minutes.
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That’s the amount of time that would take the plane to go 300 kilometers north.
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Let’s now summarize what we’ve learned about velocity vectors.
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An object’s velocity vector gives us two pieces of information.
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It tells us how fast the object is moving and in what direction.
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Second, an object’s speed is equal to the magnitude of its velocity.
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Speed, 𝑠, is equal to magnitude of 𝑣 which for a three-dimensional velocity equals the square root of 𝑣 sub 𝑥 squared plus 𝑣 sub 𝑦 squared plus 𝑣 sub 𝑧 squared.
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And, finally, velocity vectors can be added graphically, putting vectors tip to tail with one another or algebraically by adding their numerical components.