B moves with vB1 = + 1.0 m/s. During the second part of the problem, person A moves with vA2 = – 5.0 m/s while person B still moves with vB2 = + 1.0 m/s. Image Added Note |
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Can you think of a reason that it might have been a good idea to put x = 0 m at the summit instead of the parking lot? We will encounter one such reason at the end of this method. | Mathematical Representation For our chosen model, there is only one Law of Change: Latex |
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\begin{large} \[ x = x_{\rm i} + vt\]\end{large} |
Because we have divided the problem, however, we must apply this one law a total of four times (person A during part 1, person B during part 1, person A during part 2, and person B during part 2). Thus, we have: Latex |
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\begin{large} \[ x_{\rm A1}=x_{\rm A1,i}+v_{\rm A1}t_{1}\]\[ x_{\rm B1}=x_{\rm B1,i}+v_{\rm B1}t_{1}\]\[ x_{\rm A2}=x_{\rm A2,i}+v_{\rm A2}t_{2}\]\[ x_{\rm B2}=x_{\rm B2,i}+v_{\rm B2}t_{2}\] \end{large} |
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It is important that although persons A and B will potentially have different positions and velocities during each part, they share the same time. When part 1 ends, the same amount of time has elapsed for each person. Thus, in the equations, t is only labeled with "1" or "2", not "A1" or "A2". | Split the Problem – Runner on the way up Four equations seems intimidating, but in this case a systematic approach gives quick results. Begin with part 1. In part 1, both persons start in the parking lot, meaning that (in our coordinate system) x1A,i = x1B,i = 0 m. This means: Latex |
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\begin{large} \[x_{\rm A1} = v_{\rm A1}t_{1}\]\[x_{\rm B1}=v_{\rm B1}t_{1}\]\end{large} |
We have already determined the velocities, but we still have two unknowns in each equation. To solve either one, we must remember how we defined the parts of our problem. If we recall that part 1 ends when person A reaches the summit, then we must have xA1 = 3000 m. We can use this to solve for t1: Latex |
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\begin{large} \[t_{1} = \frac{x_{\rm A1}}{v_{\rm A1}} = 600 \:{\rm s}\] \end{large} |
Now, because t1 is in person B's equation also, we find: Latex |
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\begin{large} \[ x_{\rm B1} = x_{\rm A1} \frac{v_{\rm B1}}{v_{\rm A1}} = 600 \:{\rm m}\] \end{large} |
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If you are evaluating each expression as you go, think about whether the numbers make sense. Given that person B walks at 1 m/s, we certainly do expect that t1 and xB1 will be the same. | Split the Problem – Runner on the way back With part 1 understood, we move on to part 2. The first important realization here is that part 2 ends when the two persons meet. Thus, we must have xA2 = xB2. From our four original equations, that means: Latex |
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\begin{large} \[x_{\rm A2,i}+v_{\rm A2} t_{2} = x_{\rm B2,i}+v_{\rm B2}t_{2}\]\end{large} |
The second important realization is that part 2 begins where part 1 ends. This relationship is expressed mathematically by writing: Latex |
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\begin{large} \[x_{\rm A2,i} = x_{\rm A1} \]\[x_{\rm B2,i} = x_{\rm B1} = x_{\rm A1}\frac{v_{\rm B1}}{v_{\rm A1}}\]\end{large} |
which means: Latex |
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\begin{large}\[t_{2} = \frac{x_{\rm A1} \left(1-\frac{v_{\rm B1}}{v_{\rm A1}}\right)}{v_{\rm B2} - v_{\rm A2}} = \frac{x_{\rm A1}}{v_{\rm A1}}\frac{v_{\rm A1} - v_{\rm B1}}{v_{\rm B2} - v_{\rm A2}} = 400\:{\rm s}\]\end{large} |
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Notice that vB2 - vA2 is 6.0 m/s, not -4.0 m/s, since vA2 is negative 5.0 m/s. It is easy to make mistakes with negative signs. In this case, however, your final answer for the meeting location will clearly indicate there has been a mistake if you subtract incorrectly. (Try it and see what happens.) |
With t2 in hand, we know the time of the meeting. We still do not know the position, however. To get the position, we have to substitute our answer for t2 into either the equation for xA2 or that for xB2. Selecting the equation for person A: Latex |
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\begin{large}\[ x_{\rm A2} = x_{\rm A1} + v_{\rm A2}\frac{x_{\rm A1}}{v_{\rm A1}}\frac{v_{\rm A1}-v_{\rm B1}}{v_{\rm B2}-v_{\rm A2}} \]\end{large} |
A complicated equation like this is worth simplifying to see if we can make some sense of it. Some algebra will enable us to get rid of the compound fractions: Latex |
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\begin{large} \[ x_{\rm A2} = x_{\rm A1} \left(\frac{v_{\rm B2}-v_{\rm A2} + \frac{v_{\rm A2}}{v_{\rm A1}}(v_{\rm A1}-v_{\rm B1})}{v_{\rm B2}-v_{\rm A2}}\right) = \frac{x_{\rm A1}}{v_{\rm A1}}\left(\frac{v_{\rm B2}v_{\rm A1}-v_{\rm A2}v_{\rm B1}}{v_{\rm B2}-v_{\rm A2}}\right)\]\end{large} |
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Some checks of this expression are possible. Substituting t2 into the expression for xB2 should give the same answer. If the walker's speed is zero or the runner's speed is infinite, the meeting should take place in the parking lot (x=0). |
It is important to recognize that we are not finished yet. The problem asks for the distance of the meeting from the summit. We have found the position of the meeting. To find the distance requested, we must calculate: Latex |
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\begin{large} \[ |x_{\rm summit} - x_{\rm meeting}| = x_{\rm A1} - \frac{x_{\rm A1}}{v_{\rm A1}}\left(\frac{v_{\rm B2}v_{\rm A1}-v_{\rm A2}v_{\rm B1}}{v_{\rm B2}-v_{\rm A2}}\right) = \frac{x_{\rm A1}}{v_{\rm A1}} \frac{v_{\rm A2}v_{\rm B1}-v_{\rm A1}v_{\rm A2}}{v_{\rm B2}-v_{\rm A2}} = 2000 \;{\rm m} \]\end{large} |
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This equation too can be checked. Now the limit as the walker's speed goes to zero or the runner's to infinity should be x = 3000 m. Also, once you have calculated the value, it is a good idea to check that the runner has traveled 5 times as far as the walker (is that true for our answer?). |
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