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In this video, we will learn how to distinguish between velocity and speed and how to solve problems involving average velocity and average speed.
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Speed is a measure of how fast an object moves.
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For example, if we’re driving in a car, the speedometer will tell us how fast the car is going at any given time.
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It’s a rate at which the car covers a certain distance.
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Speed is a scalar quantity.
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And it’s measured as a distance per unit of time, for example, kilometers per hour.
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However, because it’s a scalar quantity, the speed doesn’t give us the direction in which the object travels.
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If we wanted the direction, then we’d need to use velocity because velocity is a vector quantity that gives us both the speed and direction of travel.
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For example, if a car is traveling at 80 kilometers per hour, its speed is 80 kilometers per hour.
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However, if we say that the car travels in a northeasterly direction at 80 kilometers per hour, that’s the velocity, since we have the speed and the direction of travel of the car.
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Let’s consider an object traveling between two points, for example, this fish.
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There are an infinite number of paths of different lengths that the object could take.
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The length of a path between two points is the distance an object travels between the points.
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For this example, it’s the fish’s distance.
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However, when we’re considering the velocity of an object, we consider its displacement.
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The displacement measures how far and in what direction the second point is from the first.
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Let’s consider another example, this time of a bird in flight.
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Let’s say that it flies 30 kilometers due east and then 40 kilometers due north.
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If we wanted to work out the total distance covered by the bird, then we would add together the distances in each of the two stages.
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So 𝐴 to 𝐵 is 30 kilometers, and 𝐵 to 𝐶 is 40 kilometers, giving us a distance of 70 kilometers.
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If instead we wanted to work out the displacement, then that’s the directed line segment between 𝐴 and 𝐶.
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We can work out the magnitude of this line segment by recognizing that 𝐴𝐵𝐶 is a right triangle.
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And we could then apply the Pythagorean theorem.
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This would give us the magnitude of the displacement as 50 kilometers.
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In order to work out the direction, we could take this angle 𝐵𝐴𝐶 as 𝜃 degrees.
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We could then note that tan of this angle 𝜃 is 40 over 30.
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And so 𝜃 must be 53.130 degrees.
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We can then convert this into degrees, minutes, and seconds to give us that the bird’s displacement is 50 kilometers, 53 degrees, seven minutes, and 48 seconds in the northeasterly direction.
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Notice that this displacement has both magnitude and a direction.
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We can then make some notes about speed and velocity.
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Speed, which is a scalar quantity, is the rate at which an object covers distance.
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It’s calculated as speed is equal to distance over time.
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Velocity, however, is a vector quantity.
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And it’s the magnitude and direction of an object’s change in position.
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It’s calculated as velocity is equal to displacement over time.
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Often, however, we need to calculate average speed.
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And that’s equal to the total distance covered over the total time taken.
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If we wish to calculate the average velocity, then that’s equal to the net displacement divided by the total time.
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The net displacement is the displacement measured directly from the object’s starting position to its end position.
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Since displacement is a vector quantity, then average velocity is also a vector quantity.
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And it can be positive or negative.
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However, the magnitude, which is scalar, is measured as a distance per unit of time.
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And it’s always positive.
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We’ll now see how we can apply these formulas in the following examples.
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An object moves north at 12 meters per second for 10 seconds and then stops and stays motionless for 10 seconds before moving north at 12 meters per seconds for another 10 seconds.
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What is the object’s average northward velocity?
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Let’s consider the three different stages of this object’s movement.
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In the first stage, the object moves north at 12 meters per second for 10 seconds.
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We’re then told that the object stops and stays motionless for 10 seconds.
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And then, finally, it moves north at 12 meters per second for another 10 seconds.
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We’re then asked to calculate the average northward velocity.
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So we can recall that average velocity is equal to the net displacement divided by the total time.
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Because we have no change in the direction, then the magnitude of this net displacement will be the same as the total distance traveled.
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We can then work out the distance in each stage.
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We can remember that distance is calculated by speed times time.
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So in the first stage, we have a speed of 12 and a time of 10 seconds.
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Multiplying those together would give us a distance of 120 meters.
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In the second stage, when the object is at rest, the speed is zero and the time is 10 seconds.
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So the distance traveled will be zero meters.
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Finally, we have that distance in the third stage is the same as the first stage.
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It’s 12 times 10, which is 120 meters.
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When we add these three values together then, we get that the total distance is 240 meters.
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In order to apply the formula for average velocity, we’ll need to calculate the total time.
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So we have 10 seconds, 10 seconds, and 10 seconds, which gives us a total time of 30 seconds.
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We can then fill these values in to the formula, remembering that we can use the distance in this case because there’s no change in direction here.
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So the magnitude of this displacement is the same as the distance traveled.
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So 240 over 30 is equal to eight.
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And the units here will be meters per second.
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And so we can give the answer for the average northward velocity as eight meters per second.
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One other way we could’ve approached this problem is by using a displacement–time graph.
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In the first stage, when the object moved for 10 seconds, remember that we calculated that displacement as 120 meters.
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It then stayed still for 10 seconds.
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And finally, it moved another 12 meters per second for 10 seconds.
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In the first and third sections, we have a positive motion.
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And so the graph has a positive slope.
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In the middle section, this was a rest.
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And so there’s zero slope.
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If we want to calculate the velocity at any stage, we can find the slope or gradient of that section.
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If we want to find the average velocity, then we can create a line segment between the starting coordinate and the end coordinate.
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We can remember that if we’ve got two coordinates 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two, we can calculate the slope as 𝑦 two minus 𝑦 one over 𝑥 two minus 𝑥 one.
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So for the two coordinates then 30, 240 and zero, zero, the slope would be calculated as 240 minus zero over 30 minus zero.
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And this simplifies to 240 over 30, which is eight meters per second.
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And so we have confirmed the original answer using a displacement–time graph.
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Let’s look at another example.
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A person is late for an appointment at an office that is at the other end of a long, straight road to his home.
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He leaves his house and runs towards 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 his appointment.
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He runs back home at the same speed he ran at before and spends 185 seconds looking for the documents, and then he runs towards his appointment again.
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This time, he runs at 5.5 meters per second for 260 seconds and then arrives at the office.
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We’re then asked three different questions.
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So let’s begin with the first one.
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How much time passes between the person first leaving his house and arriving at his appointment?
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It might be helpful to begin by visualizing what happens at each stage of this person’s journey.
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The journey begins with running for 45 seconds towards the office.
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The person then realizes that they’ve forgotten something they need, so they go home at the same speed as they traveled before.
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So that means that the time will also be 45 seconds.
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They then spend 185 seconds looking for these documents but not traveling anywhere.
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And then, finally, he runs towards the office at 5.5 meters per second for 260 seconds.
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We’re then asked for the time that passes between first leaving and then arriving at the appointment.
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So that means that we just add up the four time periods: 45 seconds, 45 seconds, 185 seconds, and 260 seconds.
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And when we work that out, we get 535 seconds.
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And that’s the answer for the first part of this question.
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The next question asks us, what is the distance between the person’s house and his office?
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In order to find the distance, we can use this information on the last stage of the journey, when we’re given the speed and the time taken.
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We can remember that distance is equal to speed multiplied by time.
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So we can fill in the values then.
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The speed is 5.5, and the time is 260.
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It is always worthwhile making sure that we do have the same equivalent units.
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In each case, the time unit is given in seconds, so we can simply multiply these values.
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When we work this out, we get a value of 1430.
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And the units here will be the distance units of meters.
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And that’s the second part of this question answered.
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The third part of this question asks, what is the person’s average velocity between first leaving his house and finally arriving at his office?
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Give your answer to two decimal places.
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We can recall the formula that average velocity is equal to net displacement over total time.
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In this problem, the net displacement will simply be the direct distance between the man’s home and the office.
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We have already calculated this distance in the second part of the question.
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It’s 1430.
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And the total time taken in the whole journey was 535 seconds.
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This gives us 2.672 and so on.
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And when we round that to two decimal places, we have a value of 2.67 meters per second.
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So if the positive direction is from home towards the office, then the person’s average velocity can be given as 2.67 meters per second.
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It’s worth noting that if we’d been asked for the average speed instead, we would’ve needed to know the distances in the first two parts of the journey along with the distance in the final part of the journey.
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In this case, average speed would have been calculated by the total distance divided by the total time.
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However, since average velocity uses displacement, then we have the value of 2.67 meters per second.
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In the final question, we’ll see an example where we find the average velocity when there’s a change in direction of motion.
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A man walked six kilometers in an easterly direction for 1.2 hours.
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He then walked eight kilometers in a northerly direction for two hours.
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Calculate the magnitude of the average velocity of the man.
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Let’s begin by thinking about this man’s journey.
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To do this, we’ll need to be familiar with our compass directions.
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In the first part of the journey then, he walks six kilometers in an easterly direction.
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And he does that for 1.2 hours.
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He then walks eight kilometers in a northerly direction, or north, for two hours.
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In order to calculate the average velocity, we can recall the formula that average velocity is equal to net displacement over total time.
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We can consider that the total time in this case is relatively simple to calculate; it’s simply 1.2 plus two, which is 3.2 hours.
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However, we need to consider what the net displacement is.
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Well, the net displacement is the displacement between the starting and final positions.
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We know that the man traveled six kilometers easterly and eight kilometers in a northerly direction.
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In order to calculate this displacement, we can observe that we have a right triangle.
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And so we could apply the Pythagorean theorem.
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Therefore, the net displacement is equal to the square root of six squared plus eight squared, which is the square root of 100.
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And that’s 10 kilometers.
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And you may have already observed that this is, of course, a three-four-five Pythagorean triple.
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It’s worth noting that usually as displacement is a vector quantity, then we would need to specify a direction of displacement.
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We could do this by calculating the angle between the horizontal and the direction of displacement.
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Or we could even note that the displacement is in an approximately northeasterly direction.
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However, since we’re asked simply for the magnitude of the average velocity, then we don’t need to do this here.
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And so we can simply fill in the value for the net displacement into the formula to calculate the average velocity.
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10 divided by 3.2 gives us 3.125, which we can give to one decimal place as 3.1.
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The units here will be kilometers per hour as the displacement is in kilometers and the time is in hours.
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And so we can give the answer that the magnitude of the average velocity is 3.1 kilometers per hour.
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We can now summarize the key points of this video.
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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.
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Speed is calculated by distance divided by time.
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Speed is measured as a distance per unit of time.
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Velocity is a vector quantity and specifies both the magnitude and the direction of an object’s change in position.
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We calculate velocity by dividing the displacement by the time.
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Velocity is given as a distance per unit of time and has a direction.
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We saw the two formulas for the averages.
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The average speed is given as total distance divided by total time.
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And the average velocity is given as net displacement divided by total time.
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Finally, we saw that 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 starting and finishing positions.