An observer stands at a point P, one unit away from a track. Two runners start at the point S in the figure and run along the track. One runner runs 2 times as fast as the other. Find the maximum value of the observer's angle of sight è between the runners.
Thank you so much!
To find the maximum value of the observer's angle of sight, we can consider the scenario when the faster runner is directly in line with the observer at point P. Let's call this point Q.
Now, let's consider the triangle PQS, where Q is directly in line with P and S. The distance between P and Q is one unit, and the distance between Q and S is also one unit.
Since the faster runner runs 2 times as fast as the other, the ratio of their speeds can be represented as 2:1. Let's call the speed of the slower runner x units per second. Therefore, the speed of the faster runner is 2x units per second.
Let's consider the time it takes for the faster runner to reach point Q. Since the distance between S and Q is one unit, and the speed of the faster runner is 2x units per second, the time taken by the faster runner is 1/(2x) seconds.
Now, let's consider the distance covered by the slower runner during this time. The distance covered by the slower runner is given by the speed multiplied by the time, which is x * (1/(2x)) = 1/2 units.
Therefore, at this point, the slower runner has covered a distance of 1/2 units, and the faster runner has covered a distance of 1 unit. This means that the distance between the slower runner and Q is 1/2 units.
Now, we can calculate the angle between the observer's line of sight and the line connecting the observer to the slower runner. We can use the tangent function to calculate this angle.
tan(θ) = Opposite/Adjacent = (1/2) / 1 = 1/2
Therefore, the angle θ is the arctan(1/2).
Using a calculator or the properties of the arctan function, we find that θ ≈ 26.565 degrees.
Thus, the maximum value of the observer's angle of sight è between the runners is approximately 26.565 degrees.
To find the maximum value of the observer's angle of sight (θ) between the runners, we need to consider the situation when the runners are farthest apart from each other.
Let's assume that the slower runner runs at a speed of v units per second. Then the faster runner must be running at a speed of 2v units per second, as given in the question.
Now, let's consider the situation after time t has passed. The slower runner will have covered a distance of vt, while the faster runner will have covered a distance of 2vt.
To find the farthest distance between the runners, we need to find the distance between their positions. This can be done by considering the right-angled triangle formed by the observer, the slower runner, and the faster runner. The hypotenuse of this triangle represents the distance between the runners.
Since the observer is one unit away from the track (point P), the distance between the observer and the slower runner can be expressed as √((vt)^2 + 1), using the Pythagorean theorem.
Similarly, the distance between the observer and the faster runner can be expressed as √((2vt)^2 + 1).
Now, to find the maximum value of θ, we need to maximize the angle between the hypotenuse and the adjacent side of the triangle.
Using trigonometry, the tangent of θ can be defined as the ratio of the opposite side (the distance between the runners) to the adjacent side (the distance between the observer and the slower runner). Therefore, we have:
tan(θ) = (distance between runners) / (distance between observer and slower runner)
= √((2vt)^2 + 1) / √((vt)^2 + 1)
To maximize tan(θ), we need to find the maximum value of the expression (√((2vt)^2 + 1)) / (√((vt)^2 + 1)). To simplify this, we can square the expression and maximize it.
Simplifying and maximizing (√((2vt)^2 + 1))^2 / (√((vt)^2 + 1))^2, we get:
(4v^2t^2 + 1) / (v^2t^2 + 1).
To find the maximum value of θ, we need to find the maximum value of this expression. Since both the numerator and denominator are positive, the maximum value occurs when (4v^2t^2 + 1) is maximized.
Differentiating (4v^2t^2 + 1) with respect to t, we get 8v^2t = 0, which implies t = 0. Therefore, the maximum value of the expression occurs when t = 0.
Plugging in t = 0 into the expression, we get:
(4v^2(0)^2 + 1) / (v^2(0)^2 + 1) = (4v^2 + 1) / 1 = 4v^2 + 1.
Thus, the maximum value of tan(θ) is 4v^2 + 1.
Finally, to find the maximum value of θ, we can take the arctan of (4v^2 + 1):
θ = arctan(4v^2 + 1)
Therefore, the maximum value of the observer's angle of sight θ between the runners is arctan(4v^2 + 1).
well, naturally, the maximum angle will be when the two runners are at opposite ends of a diameter perpendicular to the line from the observer to the center of the circle.
If the track has radius r, then
max θ = 2arctan(r)
For lack of a diagram, I am unable to incorporate any further constraints.