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 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.
NEED COORD SYSTEM
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}} = 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 answer will clearly indicate a problem if you subtract incorrectly. (Try it and see what happens.)