Question Video: Finding the Average Velocity of a Body given Its Position Coordinates at Different Times | Nagwa Question Video: Finding the Average Velocity of a Body given Its Position Coordinates at Different Times | Nagwa

# Question Video: Finding the Average Velocity of a Body given Its Position Coordinates at Different Times

A moving particle reached a point 𝐴(−3, 8) after 2 seconds and another point 𝐵(5, 0) after 10 seconds. Find the magnitude of the average velocity of the particle 𝑣 and the direction of the average velocity 𝜃.

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### Video Transcript

A moving particle reached a point 𝐴: negative three, eight after two seconds and another point 𝐵: five, zero after 10 seconds. Find the magnitude of the average velocity of the particle 𝑣 and the direction of the average velocity 𝜃.

Let’s begin by sketching this out so we can see what’s happening. The particle reaches the point with coordinates negative three, eight after two seconds. Then, after 10 seconds, it reaches the point with coordinates five, zero as shown. We can therefore say that the motion of the particle looks a little something like this. And then the question is asking us to calculate two values. We want the magnitude of the average velocity, which is 𝑣, and the direction of the average velocity. So let’s think about the magnitude first.

Velocity is a vector quantity. It has a direction and a magnitude. When we think about the magnitude or the size of the velocity, we’re actually saying, well, that’s the speed of the object. And so to find the magnitude of the average velocity, we’re going to find the average speed using the formula speed is equal to distance divided by time. But what is the distance that the particle has traveled? Well, we find the distance by considering the magnitude of its displacement, in other words, the length of the yellow line we draw in at the beginning.

We can use the Pythagorean theorem or the distance formula to calculate this measurement. This tells us that the distance between two points is the square root of 𝑥 two minus 𝑥 one squared plus 𝑦 two minus 𝑦 one squared. If we define point 𝐴 as being 𝑥 one, 𝑦 one, then 𝑥 one is equal to negative three and 𝑦 one is equal to eight. Similarly, we’ll define the point with coordinates 𝑥 two, 𝑦 two as being five, zero. Now it wouldn’t have mattered had we chosen these values the other way round. We’d still would’ve got the same answer. And so the distance between points 𝐴 and 𝐵 is the square root of five minus negative three squared plus zero minus eight squared. That’s the square root of eight squared plus eight squared. And since eight squared is 64, the square root of eight squared plus eight squared is the square root of 128.

Now, in fact, we can simplify this at this stage. 128 is equal to 64 times two. And so the square root of 128 is the square root of 64 times the square root of two. But of course, since the square root of 64 is eight, we can say that this is in turn equal to eight root two. And the distance must be eight root two units. And so we’re now ready to work out the average speed. We have the distance, and we know that the time it took is the difference between two and 10 seconds. Well, 10 minus two is equal to eight. And so speed, which is distance divided by time, is eight root two divided by eight, which is simply root two. We actually don’t know the units here, so we’re going to say that this is a square root of two length units per second. The question defined this as being equal to 𝑣, and so we finished the first part.

The next part of this question wants us to find the direction of the average velocity. Now, the direction of the average velocity is just the direction in which the particle is moving. Now, when we consider the direction of a vector, we consider this relative to the positive 𝑥-axis, and we move in a counterclockwise direction. So we add a line parallel to the 𝑥-axis from the point 𝐴. And we choose this point because we’re considering this to be the initial point, the starting point of our particle. By extending the green line and adding a right-angled triangle, we’re going to begin by finding the angle that I’ve labeled 𝛼.

Now we already know the length of the yellow line. Remember, that was the total distance traveled, and we calculated that to be eight root two units. We can find the length of any of the other sides by either using change in 𝑦 or change in 𝑥. Change in 𝑦 gives us the length of the opposite side, and that’s eight minus zero, which is, of course, simply eight. And so we can use the sine ratio to find the value of 𝛼. The general formula says that sin of 𝜃 is opposite over hypotenuse. So, in this case, we can say sin of 𝛼 is eight over eight root two.

Notice that we’ve chosen 𝛼 rather than 𝜃 because the question tells us that the direction which we know is the reflex angle is 𝜃. Then we’re going to divide both the numerator and denominator of this fraction by eight. And so we see that sin of 𝛼 is one over root two. But, of course, we should know that sin of 45 is one over root two. So 𝛼 must be equal to 45 degrees. Since angles around a point sum to 360 degrees, we’re going to subtract 45 from 360 to find the angle that we’re interested in. 360 minus 45 is 315, and so we can say that 𝜃 is 315 degrees.