Suppose you are in a room that is a*10 meters. You can walk normally along the "a" meter long wall at "s" meters per second and crab walk in any direction at 1 meter per second. You start at one corner of the room, walk normally along the wall of length "a" meters for some distance, and then crab walk straight to the corner opposite from where you started.
Suppose that the fastest route to the opposite corner of the room is to directly crab walk from one corner of the room to the other (i.e. you walk normally for 0 distance).
What can be said about the relationship between a and s?
If you walk a distance x, then the time t to reach the opposite corner is
t(x) = x/s + √((a-x)^2 + 10^2)
Now, just find dt/dx. The fastest time is when dt/dx=0, so plug in the fact that it is when x=0. Then you will get an equation is a and s.
To determine the relationship between "a" and "s" in this scenario, we can use the concept of time.
Let's assume that the distance you walk normally along the wall of length "a" meters is "x" meters.
When you walk normally, the time taken is given by the formula: time = distance / speed. So, the time taken to walk "x" distance normally is x / s.
When you crab walk straight to the opposite corner, you are covering the remaining distance of the room, which is (a - x) meters.
Since the crab walk speed is 1 meter per second, the time taken to crab walk straight to the opposite corner is (a - x) / 1 = a - x seconds.
Now, we know that the fastest route is to directly crab walk from one corner to the other, which means the time taken when you walk normally is equal to the time taken when you crab walk straight to the opposite corner.
Therefore, we have the equation: x / s = a - x.
Simplifying this equation, we get: x = as / (s + 1).
So, the relationship between "a" and "s" is given by: x = as / (s + 1).
This equation shows that the distance you walk normally, "x", is directly proportional to both "a" and "s", but inversely proportional to (s + 1).