Find the Instantaneous Velocity of D(x) = x + 2cos(x) in (0,2π)

[indigo1]

Homework Statement

if d(x)=x+2cosx represents the distance traveled by a particle from x=0 to x= 2pi.Find an x value in (0,2pi) for which the instantaneous velocity equals the average velocity.

The Attempt at a Solution

This question is a direct application of the mean value theorem (LMVT)
D(x) = x + 2cos(x)
∴ D(0) = 0 + 2cos(0) = 2
∴ D(2pi) = 2π + 2cos(2π) = 2(π + 1)
∴ Average velocity = Total distance/Total time
= [D(2π) - D(0)]/(2π - 0)
= (2 + 2π - 2)/(2π)
= 1 m/s
Instantaneous velocity =D ' (x)
∴ v(t) = D'(x) = 1 - 2sin(x)
Given that, to find 'c' such that D ' (c) = avg velocity = 1
∴ 1 - 2sin(c) = 1
∴ sin(c) = 0
∴ c = 0 or π
But 'c' needs to lie in (0,2π)
∴ At c = π, the average velocity = instantanoeus velocity = 1

ive been at this and I am just wondering if it's right?

 

[tiny-tim]

hi indigo1!

if d(x)=x+2cosx represents the distance traveled by a particle from x=0 to x= 2pi.Find an x value in (0,2pi) for which the instantaneous velocity equals the average velocity.

…
∴ Average velocity = Total distance/Total time
= [D(2π) - D(0)]/(2π - 0)
= (2 + 2π - 2)/(2π)
= 1 m/s
Instantaneous velocity =D ' (x)
∴ v(t) = D'(x) = 1 - 2sin(x)
Given that, to find 'c' such that D ' (c) = avg velocity = 1
∴ 1 - 2sin(c) = 1
∴ sin(c) = 0
∴ c = 0 or π
But 'c' needs to lie in (0,2π)
∴ At c = π, the average velocity = instantanoeus velocity = 1

looks ok