Question

A runner 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.)

Answer 1

Answer:

speed = 6.78 m/s

Explanation:

Given Data

Radius=120 m

Constant Speed=7 m/s

Distance=240 m

To find

Speed=?

Solution

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

Use the law of cosines

x² = 120² + 240² - (2 x 120 x 240 x Cosθ ............eq(1)

differentiate with respect to t

2x (dx/dt) = - 57600 x (-Sinθ) dθ/dt..............eq(2)

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

240² = 120² + 24062 - 57600 Cosθ

to find angle

Cosθ = 1/4

From the rule of right angled triangle

Sinθ = √15/4

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

2x (dx/dt) = - 57600 x (-Sinθ) dθ/dt

(dx/dt)=(7√15/4)

as we know that speed is first derivative with respect to time as acceleration is second derivative with respect to time  

So

speed = 6.78 m/s

Answer 2

Answer:

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

$dx/dt=\frac{28800}{x}Sin\theta\times \omega$   . ........  (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

$dx/dt=\frac{28800}{240}\times \frac{7}{120}\times \frac{\sqrt{15}}{4}$

$dx/dt=\frac{7\sqrt{15}}{4}$ m/s = 6.78 m/s