Arunner sprints around a circular track of radius 120 m at a constant speed of 7 m/s. the runner's friend is standing at a distance 240 m from the center of the track. how fast is the distance between the friends changing when the distance between them is 240 m? (round your answer to two decimal places.)

6.78 m/s

Explanation:

Let the distance between the two friends is x .

Radius, r = 120 m

v = 7 m/s

Angular velocity, ω = v / r = 7 / 120 rad/s

Use the law of cosines for the triangle OAB

x^2 = 120^2 + 240^2 - 2 x 120 x 240 x Cosθ      ...... (1)

differentiate with respect to t on both the sides

2 x dx/dt = - 57600 x (-Sinθ) dθ/dt

  . ........  (2)

Put x = 240 m in equation (1), we get

240^2 = 120^2 + 24062 - 57600 Cosθ

Cosθ = 1/4

So, by the use of right angled triangle, Sinθ = √15/4

Put all the values in equation (2), we get

m/s = 6.78 m/s

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2

you can cut your hand off if your not