#### To determine

**To find:** The rate of change of the distance between runner and his friend standing at

200 m from the center of the track.

#### Answer

The rate of change of the distance between them is dldt≈6.78 m/s.

#### Explanation

**Given:**

A runner sprints around a circular track of radius 100 m at a constant speed 7 m/s. And the runner`s is standing at a 200 m from the centre of the track.

**Formula used:**

(1) Chain rule: dydx=dydu⋅dudx

**Calculation:**

Let us assume that *O* be the center of the circular track of radius 100 m and *P* be the position of the runner on the circular track and *A* be the position of his friend from the 200 m from the center of the circular track.

Let θ be the angle at the center to the runner from the horizontal and l be the distance between the runner and his friend as shown in the Figure 1 given below.

Obtain dldt when l=200 m

Here, D is the arc length of the circular track, then

D=r×θ

Substitutes *r* =100.

D=100θθ=D100

Differentiate θ with respect to the time *t.*

dθdt=ddt(D100)=1100(dDdt)

Now from the cosine rule in the ΔOPA.

l2=1002+2002−(2×100×200cosθ)

Differentiate l2=1002+2002−(2×100×200cosθ) with respect to the time *t*.

ddt[l2]=ddt[1002+2002−(2×100 00cosθ)]2ldldt=−40,000(−sinθ)dθdt [∵dydx=dydu⋅dudx]dldt=20,000l(sinθ)dθdtdldt=20,000l(sinθ)(1100⋅dDdt) [∵dθdt=1100⋅dDdt]

First find the value of sinθ.

When distance between the runner and his friend is 200 m

Substitute l=200 in l2=1002+2002−(2×100×200cosθ).

2002=1002+2002−(2×100×200cosθ)40,000cosθ=10,000cosθ=10,00040,000cosθ=14

Then,

sinθ=1−cos2θ=1−(14)2=1−116=1516

On further simplification,

sinθ=1415

Substitute l=200 and sinθ=154 and dDdt=7 in dldt.

dldt=20,000200(154)(1100×7)=7154≈6.78 m/s

Therefore, the rate of change of the distance between them is dldt≈6.78 m/s.