What way should I solve this?

Question

A running coach is watching his star sprinter at a linear speed of 8 meters per second around a circular track. The coach is OUTSIDE the track, a total of 65 meters from the center. How fast is the distance between the coach and the runner increasing when she is 34 meters from her coach (and moving away from him). Assume she "hugs" one of the lanes so her path forms a perfect circle, and that he lane is 50 meters form the center

Answer 1

In this problem, we have a circle in which the runner is running (I wish I could attach a picture here).  So we have a triangle here,

1.One side of the triangle is the distance between the runner and the center of the track, which is constant at 50 m.

2.The 2nd side of the triangle is the distance between the coach and the center of the track, which is also const. at 65m.

3.The 3rd side of the triangle is the distance between the runner and the coach, this is not constant and is changing. And we have to find the rate at which this distance is changing when it is 34 m. Let us call this distance x m.

 

By the law of cosine:

 

x² = 50² + 65² - (2.50.65.cos θ)                                                                                                  (1)

x² = 6725 - (6500*cos θ)                                                                                                            (2)

 

θ is the angle between: the line that is formed by the runner's position on the track to the center of the circle and the line that is formed by the coach's position and the center of the track.

 

So, cos θ = (6725 - x²)/6500                                                                                                        (3)

 

When x = 34 m (condition of the problem),

 

cos θ (at x=34m) = (6725 -  34²)/6500 = 0.86

 

Differentiating eq. 2 w.r.t time:

2x dx/dt = 6500 sin θ * dθ/dt                                                                                                         (4)

 

Now, sin θ (at x=34 m) = sqrt ( 1 - cos² θ) =  0.51

We have to determine dθ/dt (at x=34 m)

 

Now,  L= θ*r ( L is the arc length, don't confuse with x and r is the radius of the circle in which the runner is running)

  

        dθ/dt = (dL/dt)/r

 

dL/dt is the speed of the runner. It is 8m/s. And r = 50 m

 

Thus dθ/dt (at x=34 m, actually it doesn't depend on the value of x as you can see) = 8/50 =0.16

Now going back to eq. 4 and plugging in all the values (we have them all now):

x=34 m

sin θ = 0.51

dθ/dt = 0.16

 

2*34*(dx/dt) = 6500*0.51*0.16

dx/dt = 7.8 m/s

 

It is a little complicated. If you draw that circle with the triangle in the beginning, it will be easier to understand.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

%MCEPASTEBIN%