This is our fourth and final tutorial on how things move. In tutorials six and seven, we explored how things move and change position. In Tutorial eight, we looked at how far something moves. In this tutorial, we are going to take what was introduced in the previous tutorials to explore solving a question involving distance and position. We’re not going to introduce anything new; we’re just going to put the pieces together that we’ve already discussed.
We’re going to begin with the familiar example involving a street, a bus, and a woman. As illustrated in Figure A, the front of the bus is at x and it has been moving at velocity v for t seconds. Standing on the street is a woman who is located at the rear of the bus, represented by the black triangle. The woman has one goal: to run at velocity w from the rear to the front of the bus, where upon reaching the front she will turn around and run back to the rear of the vehicle. This example should be familiar, since it closely mirrors the initial setup presented in Tutorial eight. The only assumption we’ll make is that at time 0, the rear of the bus was located at the origin, or at the zero mark.
We’re going to ask a question involving distance and position:
Where will the bus be located when the woman has completed half of her round–trip journey and how far will she have traveled?
Revisiting Tutorial eight, we know that the total round–trip distance that the woman travels is the sum of the forward intercept length and the reflected intercept length. To find these intercept lengths we need three pieces of information, the velocity of the bus, v, the velocity of the woman, w, and the length of the bus. Therein lies the first problem: the length of the bus is not given. Fortunately, this is something we can find using the translation transformation, by rewinding the clock to see where the bus was t seconds ago. As illustrated in Figure B, we reposition the bus using the transformation:
x’ = x – vt
Because the rear of the bus is at the origin at time 0, the position of the front of the bus, x’, is also the length of the bus.
Revisiting Tutorial one, we could use either x’ as the length of the bus or we could use the expression x – vt; both are equivalent. For now, we’ll use x’, but we will come back to this later.
Recall from Tutorial eight that we can answer questions about the distance the woman runs without regard to the position of the bus or the woman. So, we could leave the bus where it is, as in Figure B, with its rear at the origin. Unfortunately, doing so would place us at a disadvantage later when we need to determine the position of the bus after the woman has run the average intercept length. So, as illustrated in Figure C, we’ll move it back to its original position at time t. Notice the only real difference between figures A and C is that we now know the length of the bus.
Now that we know the length of the bus, x’, we can determine the forward and reflected intercept times and lengths. As discussed in Tutorial eight, the forward intercept length is illustrated in Figure D. It is the length from the triangle to the circle, which is longer than the length of the bus. The reflected intercept length is illustrated in Figure E. It is the length from the triangle to the circle, which in this case is shorter than the length of the bus. Notice that when each intercept length is divided by the woman’s velocity w we have the forward and reflected intercept times, respectively. Before we continue, there is an important subtlety that we must note: The woman runs in both directions. Once she has run the total length of one oscillation, she is not ahead of the bus; she is at the rear of the vehicle.
So far, we’ve done a lot of work, but haven’t answered the original question. Fortunately, we have the information we need to answer the question. Because we have the forward and reflected intercepts, we can find their average by using the addition mean equation or the subtraction mean equation. This average is illustrated in Figure F.
Notice that the average intercept length is shorter than the forward intercept length, but longer than the reflected intercept length. This means that when the woman has reached the average intercept length, she has not yet reached the front of the bus. Remember, she doesn’t reach the front of the bus until she has traveled the forward intercept length, the position of which is represented by the black circle. This is an extremely subtle, but important point to remember; one that we’ll come back to when we examine Einstein relativity theory.
Figure F also illustrates the answer to the part of the question asking how far the woman will have traveled to arrive at the halfway point of her round–trip journey. Simplifying the equation, the distance that the woman runs, or the average intercept length, ξ is:
ξ = x’ / (1 – v2 / w2 )
Don’t be confused by the Greek variable that is used to represent the average intercept length. Here we use it for a specific reason, to represent a distance. This will help to prevent mistaking a distance for a position.
Remember earlier when we said x’ was equivalent to x – vt? Now we put that information to use. Since they’re equivalent, we can replace x’ with x – vt to restate the average intercept length as:
ξ = (x –vt) / (1 – v2 / w2 )
Now that we have answered the question about the distance the woman has run, we must now answer the accompanying question of the location of the bus when she reaches this distance. Remember, she has not yet reached the front of the bus.
Fortunately, since the bus is moving forward at velocity v, all we need to know is how long she’s been in motion and we can use that amount of time in the translation transformation to find the bus’s position. The amount of time that it takes the woman to run the average intercept length is called the average intercept time, and it is found by dividing the average intercept length by w, the velocity of the woman:
τ = ξ / w
Again, we use the Greek letter to represent the amount of time that the woman takes to run half her journey to avoid confusing it with time t, which is the amount of time that the bus has already been in motion. So, if time starts at 0, then the time we are considering is t+τ. We can solve this question in one of two ways. First, we could use this total time in the translation equation using the vehicle’s position x’ at time 0:
x” = x’ + v ( t + τ )
Or we could use the amount of time that the woman was running in the translation equation (aka, average intercept time) from the vehicle’s position x at time t:
x” = x + vτ
The first approach is illustrated in Figure G, although both will give the same answer. Notice that the front of the bus is ahead of the woman. They will not meet until they both arrive at the forward intercept point.
We have now answered the question about the bus’s location and how far the woman has run when she has completed half of her journey.
Now for the good news: You have mastered almost all that you need to know about Modern Mechanics! Additionally, you now have a solid foundation from which you can explore Einstein’s relativity theory as well as see where he made several mistakes. Both of these we’ll explore in upcoming tutorials.
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