h3. Method 1
{toggle-cloak:id=diag} {color:red} *Diagrammatic Representation* {color}
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One way to be sure that each person has constant velocity is to split the problem into two parts. The point of division is when person A reaches the summit and turns around. If we set up a one-dimensional coordinate system as shown below, then during the first part of the problem person A moves with velocity _v_~A1~ = + 5.0 m/s and person B moves with _v_~B1~ = + 1.0 m/s. During the second part of the problem, person A moves with _v_~A2~ = -- 5.0 m/s while person B still moves with _v_~B2~ = + 1.0 m/s.
!meet.png!
{note}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.{note}
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{toggle-cloak:id=math} {color:red} *Mathematical Representation* {color}
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For our chosen model, there is only one Law of Change:
{latex}\begin{large} \[ x = x_{\rm i} + vt\]\end{large}{latex}
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}\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}{latex}
{note}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".{note}
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{toggle-cloak:id=up} {color:red} *Split the Problem -- Runner on the way up* {color}
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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) _x_~1A,i~ = _x_~1B,i~ = 0 m. This means:
{latex}\begin{large} \[x_{\rm A1} = v_{\rm A1}t_{1}\]\[x_{\rm B1}=v_{\rm B1}t_{1}\]\end{large}{latex}
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 _x_~A1~ = 3000 m. We can use this to solve for _t_~1~:
{latex}\begin{large} \[t_{1} = \frac{x_{\rm A1}}{v_{\rm A1}} = 600 \:{\rm s}\] \end{large} {latex}
Now, because _t_~1~ is in person B's equation also, we find:
{latex}\begin{large} \[ x_{\rm B1} = x_{\rm A1} \frac{v_{\rm B1}}{v_{\rm A1}} = 600 \:{\rm m}\] \end{large}{latex}
{tip}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 _t_~1~ and _x_~B1~ will be the same.{tip}
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{toggle-cloak:id=down} {color:red} *Split the Problem -- Runner on the way back* {color}
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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 _x_~A2~ = _x_~B2~. From our four original equations, that means:
{latex}\begin{large} \[x_{\rm A2,i}+v_{\rm A2} t_{2} = x_{\rm B2,i}+v_{\rm B2}t_{2}\]\end{large}{latex}
The second important realization is that part 2 begins where part 1 ends. This relationship is expressed mathematically by writing:
{latex}\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}{latex}
which means:
{latex} \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}{latex}
{warning}Notice that _v_~B2~ - _v_~A2~ is 6.0 m/s, *not* -4.0 m/s, since _v_~A2~ 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.){warning}
With _t_~2~ 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 _t_~2~ into either the equation for _x_~A2~ or that for _x_~B2~. Selecting the equation for person A:
{latex}\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}{latex}
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}\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}{latex}
{tip}Some checks of this expression are possible. Substituting _t_~2~ into the expression for _x_~B2~ 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).{tip}
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:{color}
{latex}\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}{latex}
{tip}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?).{tip}
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