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QUESTION: 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.

ANSWER: (a) Assuming to the right is positive, it is moving to the right for all t

except when t is in the interval (π/2, 3π/2).

(b) To find this, integrate v(t) from 0 to 2π.

The value of the integral of f(t) = 3cos(2t) is F(t).

Doing this gives ∫f(t)dt = ∫3cos(2t)dt = 3sin(2t)/2 = 1.5*sin(2t) = F(t).

The answer would be F(2π) - F(0).

(c) To find the total distance travelled, it would be done by evaluating three integrals.

Each of the integrals would be the one above, the the limits of integration on each would

be different. It has to do with knowing cos(2t) is 0 at t = π/4 and t = 3π/4.

The 1st integral has the value |F(π/2) - F(0)|,

the 2nd integral has the value |F(3π/2) - F(π/2)|, and

the 3rd integral has the value |F(2π) - F(3π/2)|.

The value of the 2nd integral would be a negative distance,

since it is moving in a negative direction at that point.

The answer would be twice this value, since the movement from 0 to π

is the same as the movement from π to 2π.

---------- FOLLOW-UP ----------

QUESTION: I am just wondering if you could explain more about part a). How is it that you determine which intervals move in the positive direction. That is the only thing i do not understand from your explanation. Everything else was crystal clear.

If the velocity were given by a cos(t) funciton,

we know cos(t) is positive in the 1st and 4th quadrants.

Since we have a cos(2t), the function would be positive in the 1st half of the 1st quadrant,

the 2nd half of the 2nd quadrant, the 1st half of thew 3rd quadrant, and the 2nd half of the fourth quadrant.

Calculus

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