The velocity function of a moving particle on a coordinate line is v(t) = 3cos(2t) for 0 is less than or equal to t
and 2pi is greater than or equal to t.
(c) Determine the total distance travelled by the particle during 0 is less than or equal to t and 2pi is greater than or equal to t.
The distance function d(t) is the integral of the velocity function v(t)
So `d(t) = int v(t) dt = int 3cos(t) dt = 3/2 sin(2t)` + C
because `d/dt sin(2t) = 2cos(2t)` and the 2's cancel
The distance function looks like this
The particle moves distance 1.5 units forwards in time t = `pi/4` then another 3 units backwards until `t = 3/4 pi` then another 3 units forwards until `t = 5/4 pi` then another 3 units backwards until `t = 7/4 pi` . Finally it moves distance 1.5 units until `t = 8/4 pi = 2pi` forwards to return to where it was. In total that is distance 1.5 + 3*3 + 1.5 = 12 units
answer: the particle moves 12 units in total but swings between -1.5 and 1.5 units from the start point