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 vA = 5.0 m/s. Person B walks the trail at a constant speed vB=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.
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.
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 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.
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.
For our chosen model, there is only one Law of Change:
\begin
[ x = x_
+ vt]\end
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:
\begin
[ x_
=x_
+v_
t_
][ x_
=x_
+v_
t_
][ x_
=x_
+v_
t_
][ x_
=x_
+v_
t_
] \end
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".
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:
\begin
[x_
= v_
t_
][x_
=v_
t_
]\end
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:
\begin
[t_
= \frac{x_{\rm A1}}{v_{\rm A1}} = 600 \:
] \end
Now, because t1 is in person B's equation also, we find:
\begin
[ x_
= x_
\frac{v_{\rm B1}}{v_{\rm A1}} = 600 \:
] \end
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.
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:
\begin
[x_
+v_
t_
= x_
+v_
t_
]\end
The second important realization is that part 2 begins where part 1 ends. This relationship is expressed mathematically by writing:
\begin
[x_
= x_
][x_
= x_
= x_
\frac{v_{\rm B1}}{v_{\rm A1}}]\end
which means:
\begin
[t_
= \frac{x_
\left(1-\frac{v_{\rm B1}}{v_{\rm A1}}\right)}{v_
- v_{\rm A2}} = \frac{x_{\rm A1}}{v_{\rm A1}}\frac{v_
- v_{\rm B1}}{v_
- v_{\rm A2}} = 400\:
]\end
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:
\begin
[ x_
= x_
+ v_
\frac{x_{\rm A1}}{v_{\rm A1}}\frac{v_
-v_{\rm B1}}{v_
-v_{\rm A2}} ]\end
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:
\begin
[ x_
= x_
\left(\frac{v_
-v_
+ \frac{v_{\rm A2}}{v_{\rm A1}}(v_
-v_
)}{v_
-v_{\rm A2}}\right) = \frac{x_{\rm A1}}{v_{\rm A1}}\left(\frac{v_
v_
-v_
v_{\rm B1}}{v_
-v_{\rm A2}}\right)]\end
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:
\begin
[ |x_
- x_
| = x_
- \frac{x_{\rm A1}}{v_{\rm A1}}\left(\frac{v_
Unknown macro: {rm B2}v_
-v_
v_{\rm B1}}{v_
-v_{\rm A2}}\right) = \frac{x_{\rm A1}}{v_{\rm A1}} \frac{v_
v_
-v_
v_{\rm A2}}{v_
-v_{\rm A2}} = 2000 \;
]\end
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.
Method 2
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 runner has come up the mountain must add to equal the distance to the summit. Again, the sketch might help you to see this.
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:
\begin
[ 2d = v_
t + v_
t ] \end