}{composition-setup}
{excerpt:hidden=true}Two people moving in one dimension with constant speed are destined to meet -- but where?{excerpt}
Two people have decided to use a mountain trail to get some exercise. They start out from the parking lot at the bottom of the trail at the same time. Person A runs the trail at a constant speed |_v_~A~| = 5.0 m/s. Person B walks the trail at a constant speed |_v_~B~|=1.0 m/s. Given that the people must return along the same path they climbed up, and given that the summit of the trail is _d_ = 3.0 km from the parking lot, how far from the summit will the people be when they meet going in opposite directions? (Assume neither person pauses.)
{toggle-cloak:id=sys} *Systems:* {cloak:id=sys}Each person will be treated as a point particle.{cloak}
{toggle-cloak:id=int} *Interactions:* {cloak:id=int}Not relevant to this model.{cloak}
{toggle-cloak:id=mod} *Model:* {cloak:id=mod}[1-D Motion (Constant Velocity)] applies to each person separately. Depending upon how you visualize the problem, the model may have to be applied twice to the runner (person A). We will suggest two possible methods by which to apply this model in the Approach.{cloak}
{toggle-cloak:id=app} *Approach:*
{cloak:id=app}
This problem stretches the definition of One-Dimensional Motion with Constant Velocity. Even if we assume the path is perfectly straight, the runner must reverse direction at the summit, and so it would seem that person A's velocity changes its mathematical sign within the problem. We will present two ways to deal with this issue. The first is more straightforward conceptually, but is slightly more tedious. The second requires deeper physical reasoning, but is slightly faster.
{info}Even though the _dynamics_ of the motions described in this problem are very different if the path is curvy instead of straight, the _kinematics_ are mathematically equivalent. It is mathematically possible to parameterize the motion along a non-self-intersecting path as a one-dimensional motion. Since this problem only deals with kinematics, our conclusions are valid for a curvy path as well.{info}
{deck:id=methdeck}
{card:label=Method 1: Split the Motion}
h3. Method 1: Split the Motion
{toggle-cloak:id=diag} {color:red} *Diagrammatic Representation* {color}
{cloak:id=diag}
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}
{cloak:diag}
{toggle-cloak:id=math} {color:red} *Mathematical Representation* {color}
{cloak:id=math}
For our chosen model, there is only one Law of Change:
{latex}| Card |
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| label | Method 1: Split the Motion |
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| 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
| 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: |
\begin{large} \[ x = x_{\rm i} + vt\]\end{large} |
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{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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{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}
{cloak:math}
{toggle-cloak:id=up} {color:red} *Split the Problem -- Runner on the way up* {color}
{cloak:id=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) _x_~1A,i~ = _x_~1B,i~ = 0 m. This means:
{latex}| 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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\begin{large} \[x_{\rm A1} = v_{\rm A1}t_{1}\]\[x_{\rm B1}=v_{\rm B1}t_{1}\]\end{large} |
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{latex}
have _x_~A1~ = 3000 m. We can use this to solve for _t_~1~:
{latex}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} |
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{latex}
__~1~
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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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{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, |
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__~1~ __~B1~ {tip}
{cloak:up}
{:=down} {color:red} *Split the Problem -- Runner on the way back* {color}
{cloak:id=down}
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}| 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} |
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{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} |
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{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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{latex}
{warning}Notice that _v_~B2~ - _v_~A2~ is | Warning |
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Notice that vB2 - vA2 is 6.0 |
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** __~A2~ ** {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}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} |
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{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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{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). |
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{tip}
{} |
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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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{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}
{cloak:down}
{card}
{card:label=Method 2: Total Distance}
h3. Method 2: Total Distance
{toggle-cloak:id=diag2} {color:red} *Diagrammatic Representation* {color}
{cloak:id=diag2}
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!
{cloak:diag2}
{toggle-cloak:id=math2} {color:red} *Mathematical Representation* {color}
{cloak:id=math2}
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}| 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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| Card |
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| label | Method 2: Total Distance |
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| 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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\begin{large} \[ 2d = |v_{\rm A}|t + |v_{\rm B}|t \] \end{large} |
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{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}| Note |
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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 |
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}{latex
{} |
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\begin{large} \[ x_{\rm B} = x_{\rm B,i} + v_{B} t \] \end{large} |
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{latex}
__~B~
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\begin{large} \[ x_{\rm B} = v_{B} \frac{2d}{|v_{\rm A}| + |v_{\rm B}|} = 1000 \:{\rm m} \] \end{large} |
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{latex}
{color}
{latex}| Latex |
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\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} |
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{latex}
{tip}Is this expression equivalent to that found in method 1?{tip}
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{card}
{deck}
{cloak:app}| Tip |
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Is this expression equivalent to that found in method 1? |
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