# 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 and 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.

Luca B. | Certified Educator

calendarEducator since 2011

starTop subjects are Math, Science, and Business

You need to remember that the total distance may be found using the absolute value function such that:

`|cos 2t| = cos 2t, 2t in (0,pi/2)U((3pi)/2,2pi)`

`|cos 2t| = - cos 2t, 2t in (pi/2,(3pi)/2)`

You need to evaluate the following definite integrals to find the total distance, such that:

`d = int_0^(pi/2) (3-cos2t) dt + int_(pi/2)^((3pi)/2) (3+cos2t) dt + int_((3pi)/2)^(2pi) (3-cos2t) dt`

Using the fundamental theorem of calculus yields:

`d = (3t-(sin2t)/2)|_0^(pi/2) + (3t+(sin2t)/2)_(pi/2)^((3pi)/2) + (3t-(sin2t)/2)|_((3pi)/2)^(2pi)`

`d = (3pi/2) - (sin pi)/2 - 0 + 0 + (9pi/2) + (sin3pi)/2 - (3pi)/2 - (sin pi)/2 + 6pi - (sin 4pi)/2 - (9pi/2) + (sin3pi)/2`

`d = 6pi`

Hence, evaluating the total distance travelled by the moving particle yields `d = 6pi.`

check Approved by eNotes Editorial