A runner sprints around a circular track of radius 100m at a constant speed of 7 m/s. The runner’s friend is standing at a distance 200m from the center of the track. How fast is the distance between the friends changing when the distance between them is 200m?
http://www2.sunysuffolk.edu/pestiej/related_rates.pdf
An object is moving along a circular path of radius r. (5 marks)
(a) Show that its velocity vector is perpendicular to the radial rays at any moment. This confirms that the velocity vector is tangent to the path, as it should be.
(b) Show that its acceleration vector is perpendicular to its velocity vector if the object rotates with a fixed speed. This confirms that the acceleration vector will be parallel to the radial ray at any moment.
Hint: Use the fact that every two vectors u = (ux,uy) and v = (vx,vy) are perpendicular exactly if uxvx + uyvy = 0.
To solve this problem, we can use the concept of rates and the Pythagorean theorem.
Let's call the distance between the runner and their friend "d" (this distance is given as 200m in the question). We want to find how fast this distance is changing when it reaches 200m.
We can consider the runner's position on the circular track and their friend as forming a right triangle. The runner is at the vertex of the right angle, the friend is at one of the other vertices, and the center of the circular track is at the third vertex.
According to the Pythagorean theorem, the distance between the friend and the center of the track (r) and the distance between the runner and the center of the track (100m) form the other two sides of the triangle.
So, we have the equation: d^2 = (100m)^2 + r^2
Now, to find how fast the distance (d) is changing, we can take the derivative of both sides of the equation with respect to time (t). This will give us:
2d * dd/dt = 2r * dr/dt
where dd/dt represents the rate at which the distance d is changing, and dr/dt represents the rate at which the distance r (the distance between the friend and the center of the track) is changing.
We are given that the runner is sprinting at a constant speed of 7 m/s, so dr/dt = 7 m/s.
Since we want to find how fast the distance d is changing when it reaches 200m, we can substitute d = 200m into the equation. We also know that r = 200m from the given information.
Substituting these values into the equation, we get:
2(200m) * dd/dt = 2(200m) * 7 m/s
Simplifying the equation, we find:
dd/dt = 7 m/s
Therefore, the distance between the friends is changing at a constant rate of 7 m/s when the distance between them is 200m.
To find the rate at which the distance between the runner and his friend is changing, we can make use of the concept of related rates.
Let's consider a right-angled triangle formed by the runner, his friend, and the center of the track. The hypotenuse of this triangle represents the distance between the runner and his friend, and the side adjacent to the right angle represents the distance from the center of the track to the friend (200m). The runner is moving along the circular track, so the distance from the center of the track to the runner remains constant at 100m.
Let's denote the distance between the runner and his friend as x. To find how fast x is changing with respect to time, we need to differentiate x with respect to time.
Using the Pythagorean theorem, we have:
x² = (200m)² + (100m)²
Differentiating implicitly with respect to time (t):
2x(dx/dt) = 2(200m)(d/dt) + 0
Simplifying:
2x(dx/dt) = 2(200m)(d/dt)
dx/dt = (200m)(d/dt) / x
Now we need to find dx/dt when x = 200m.
Plugging in the values:
dx/dt = (200m)(d/dt) / 200m
dx/dt = d/dt
Therefore, the rate at which the distance between the runner and his friend is changing is equal to the rate at which the runner is moving, which is 7 m/s. Thus, the distance between them is changing at a constant rate of 7 m/s when the distance between them is 200m.