I really have no idea how to do this. I think you are supposed to find the integral and then evaluate it but i am not sure how to proceed in doing so.
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 2pi is greater than or equal to t
(a) Determine when the particle is moving to the right.
(b) Determine when the particle stops.
(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 particle is moving to the right when 3cos(2t) > 0
since o < t < 2pi , o < 2t < 4pi
Look at a graph of cosine between 0 and 4pi .
You'll see that it's positive when 0<2t<pi/2 , 3pi/2<2t<2pi ,2pi<2t<5pi/2 , 7pi/2<2t<4pi .
So the particle is moving to the right when 0<t<pi/4 , 3pi/4<t<pi ,pi<t<5pi/4 , 7pi/4<t<2pi
The particle has stopped when 3cos(2t) = 0
Again, look at a graph of cosine betwen 0 and 4pi
cos(2t) is 0 when 2t=pi/2 , 2t=3pi/2 2t=5pi/2 , 2t =7pi/2
so the particle has stopped when
t=pi/4 , t=3pi/4 t=5pi/4 , t =7pi/4
Since the velocity at t is 3cos(2t) , integrate this over each of the intervals where the particle moves to the right
an anti derivative for 3cos(2t) is (3/2)sin(2t)
So S 3cos(2t) dt from 0 to pi/4 is (3/2)sin(pi/2) - (3/2)sin(0) = 3/2 - 0 = 3/2
S 3cos(2t) dt from 3pi/4 to pi is (3/2)sin(2pi) - (3/2)sin(3pi/2) = 0 - (-3/2) = 3/2
S 3cos(2t) dt from pi to 5pi/4 is (3/2)sin(5pi/2) - (3/2)sin(2pi) = 3/2 - 0 = 3/2
S 3cos(2t) dt from 7pi/4 to 2pi is (3/2)sin(4pi) - (3/2)sin(7pi/2) = 0 - (-3/2) = 3/2
Thus , the total distance traveled to the right is
3/2 + 3/2 + 3/2 + 3/2 = 6
Since the particle ends where it begins at 0 , it must travel to the left as far as it travels to the right. So the total distance traveled to the left is also 6
The total distance traveled is therefore 12 units .
(Note that we can't simply integrate 3cos(2t) from 0 to 2pi, because this will subtract the distances traveled to the left instead of adding and give us zero for distance traveled)