Total distance travelled from velocity function (calculus)

An object. Moves along the x-axis with initial position x(0) = 2. The velocity of the object at time t >/= 0 is given by v(t) = sin(t pi/3).
What is the total distance travelled by the object over the time interval 0

I'm looking for the non-integral way.

Thanks in advance!!!

v(t) = sin(t * (pi/3))
s(t) = (3/pi) * (-cos(t * (pi/3))) + C
s(t) = C - (3/pi) * cos(t * (pi/3))
s(0) = 2

2 = C - (3/pi) * cos(0 * (pi/3))
2 = C - (3/pi) * cos(0)
2 = C - (3/pi) * 1
2 = C - (3/pi)
2 + 3/pi = C

s(t) = 2 + (3/pi) - (3/pi) * cos(t * (pi/3))
s(t) = 2 + (3/pi) * (1 - cos(t * (pi/3)))

s(4) - s(0) =>
2 + (3/pi) * (1 - cos(4 * (pi/3))) - 2 - (3/pi) * (1 - cos(0 * pi/3)))
(3/pi) * (1 - cos(4pi/3) - 1 + cos(0))
(3/pi) * (1 - (-1/2))
(3/pi) * (1 + 1/2)
(3/pi) * (3/2)
9/(2pi)

The answer that Matticus gave is the displacement of the object in question, not the total distance. Nikolaos has found the total distance for you. TD to Matticus and TU to Nikolaos.

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v(t) describes a simple harmonic oscillation (starting from one of the endpoints) with amplitude 1 and w = pi/3 (therefore the period is T = 6)
So at t = 4 it's past half the period by 1 sec
At half the period (3 sec) it has traveled two full amplitudes, that is a distance of 2.
Then it begins traveling back, for another sec. That corresponds to a phase p = w*1 = pi/3, so the distance traveled in that second is (1 - cos(pi/3) = 1-1/2 = 1/2
So the total distance is 2 + 1/2 = 5/2