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| label | Method 1: Split the Motion |
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h3. Method 1: Split the Motion
{: Diagrammatic Representation 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 vA1 = + 5.0 m/s and person 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
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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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=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|width=700!
{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}
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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:
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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} | | Note |
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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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{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}
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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~ A1 = 3000 m. We can use this to solve for _ t1: | Latex |
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_~1~:
{latex}\begin{large} \[t_{1} = \frac{x_{\rm A1}}{v_{\rm A1}} = 600 \:{\rm s}\] \end{large} {latex}
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Now, because _t _~1~ 1 is in person B's equation also, we find:
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}\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 | | Tip |
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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 _t _~1~ 1 and _x _~B1~ B1 will be the same. {tip}
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: 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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=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}
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The second important realization is that part 2 begins where part 1 ends. This relationship is expressed mathematically by writing:
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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}{latex}
| which means:
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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}{latex}
{warning}Notice that _v_~B2~ - _v_~A2~ is | | Warning |
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Notice that vB2 - vA2 is 6.0 m/s, *not * -4.0 m/s, since _v _~A2~ 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.) |
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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{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}
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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:
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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}{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 | | Tip |
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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). {tip}
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It is important to recognize that we are not finished yet. The 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:
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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}{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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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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| label | Method 2: Total Distance |
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h3. Method 2: Total Distance
{: Diagrammatic Representation Another approach to this problem is to avoid the issue of the direction switch for the runner by thinking in terms of distance alone. The key to such a restructuring of the problem is to consider the distance covered by both persons. Sketching a graph like the one below might help with this. In the graph the red line represents the runner while the blue line represents the walker. By the time they meet, the runner has already been to the summit, covering a distance d = 3000 m in the process. At this point, we do not know how far back down the mountain the runner has made it, nor do we know how far up the mountain the walker has come before they meet. Consider, however, that the distance the runner has come down the mountain plus the distance the walker has come up the mountain must add to equal the distance to the summit. Again, the sketch might help you to see this.  Image Added
Mathematical Representation With this important realization, we can say that the total distance covered by the runner in the time from leaving the parking lot plus the total distance covered by the walker since leaving the parking lot is equal to twice the summit distance, 2 d = 6000 m. Now, we simply construct an equation that says the same thing: | Latex |
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=diag2} {color:red} *Diagrammatic Representation* {color}
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Another approach to this problem is to avoid the issue of the direction switch for the runner by thinking in terms of distance alone. The key to such a restructuring of the problem is to consider the distance covered by _both_ persons. Sketching a graph like the one below might help with this. In the graph the red line represents the runner while the blue line represents the walker. By the time they meet, the runner has already been to the summit, covering a distance _d_ = 3000 m in the process. At this point, we do not know how far back down the mountain the runner has made it, nor do we know how far up the mountain the walker has come before they meet. Consider, however, that the distance the runner has come down the mountain _plus_ the distance the walker has come up the mountain must add to equal the distance to the summit. Again, the sketch might help you to see this.
!meetgraph.png|width=650!
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{toggle-cloak:id=math2} {color:red} *Mathematical Representation* {color}
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With this important realization, we can say that the total distance covered by the runner in the time from leaving the parking lot _plus_ the total distance covered by the walker since leaving the parking lot is equal to twice the summit distance, 2 _d_ = 6000 m. Now, we simply construct an equation that says the same thing:
{latex}\begin{large} \[ 2d = |v_{\rm A}|t + |v_{\rm B}|t \] \end{large}{latex}
{note}This equation is found using the definition of [distance] and [speed]. It does not follow directly from the Laws of Change of any of our models. Going outside the usual models requires confidence with the material. If you are studying physics for the first time, you will likely find Method 1 more understandable.{note}
Solving for _t_ gives:
{latex} |
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This equation is found using the definition of distance and speed. It does not follow directly from the Laws of Change of any of our models. Going outside the usual models requires confidence with the material. If you are studying physics for the first time, you will likely find Method 1 more understandable. |
Solving for t gives: | Latex |
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\begin{large} \[ t = \frac{2d}{|v_{\rm A}|+|v_{\rm B}|} = 1000 \;{\rm s}\] \end{large}{latex}
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This equation is simple, but we have to be careful about the meaning of the result. We have not set up equations for the position of either person yet. We have to do that now. The easiest equation to use is that of the walker, who does not change direction. The Law of Change for our model tells us:
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}\begin{large} \[ x_{\rm B} = x_{\rm B,i} + v_{B} t \] \end{large}{latex}
| If we assume the same coordinate system as we defined in Method 1 (see the figure above) then the walker begins at _x _~B~B=0 and we find:
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}\begin{large} \[ x_{\rm B} = v_{B} \frac{2d}{|v_{\rm A}| + |v_{\rm B}|} = 1000 \:{\rm m} \] \end{large}{latex}
| As in method 1, we must remember that this is the position of the walker at the meeting, not the distance from the summit. To find the distance from the summit we calculate: | Latex |
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{latex}\begin{large} \[ |x_{\rm summit} - x_{\rm meeting}| = d - v_{\rm B} \frac{2d}{|v_{\rm A}| + |v_{\rm B}|} = d \frac{|v_{\rm A}|+|v_{\rm B}|-2v_{\rm B}}{|v_{\rm A}| + |v_{\rm B}|} = 2000 \:{\rm m} \] \end{large}{latex}
{tip}Is this expression equivalent to that found in method 1?{tip}
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Is this expression equivalent to that found in method 1? |
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