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Two people have decided to use of 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.)  

Systems:  Each person will be treated as a point particle.

Model:  [One-Dimensional Motion with 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.

Approach:  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}

h5. Method 1

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.

NEED COORD SYSTEM

{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}

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}

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}


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:

{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}