# The velocity function of a moving particle on a coordinate line is v(t) = 3 cos(2t) for 0 is less than or equal to t is greater than or equal to 2pi (c) Determine the total distance travelled by...

The velocity function of a moving particle on a coordinate line is v(t) = 3 cos(2t)

for 0 is less than or equal to t is greater than or equal to 2pi

(c) Determine the total distance travelled by the particle during 0 is less than or equal to t is greater than or equal to 2pi.

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to determine the total distance, hence, you need to remember that the cosine function is negative in quadrants 2 and 3 and it is positive in quadrants 1 and 4, hence, the total distance may be evaluated such that:

`v(t) = |3 cos 2t| = {(3 cos 2t, 2t in (0,pi/2)U(3pi/2,2pi)),(- 3 cos 2t, 2t in (pi/2,pi)U(pi,3pi/2)):}`

`d(t) = int_0^(pi/2) 3 cos 2t dt + int_(pi/2)^pi (- 3 cos 2t) dt + int_pi^((3pi)/2) (- 3 cos 2t) dt + int_((3pi)/2)^(2pi) 3 cos 2t dt`

`d(t) = (3/2) sin 2t|_0^(pi/2) - (3/2) sin 2t|_(pi/2)^pi - (3/2) sin 2t|_pi^((3pi)/2) + (3/2) sin 2t|_((3pi)/2)^(2pi)`

Using the fundamental theorem of calculus yields:

`d(t) = (3/2) (sin((2pi)/2) - sin 0) - (3/2)(sin 2pi - sin((2pi)/2)) - (3/2)(sin 3pi- sin pi) + (3/2)(sin 4pi - sin 3pi)`

Since `sin pi = sin 2pi = sin 3pi = sin 4pi = 0`  yields:

`d(t) = (3/2)(0 - 0 - 0 + 0 - 0 + 0 + 0 - 0)`

`d(t) = 0`

Hence, evaluating the total distance travelled by the moving particle yields `d(t) = 0.`