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Calculus/Rectilinear Motion

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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) The particle is moving to the right when it has positive velocity, ie
     3cos (2t) > 0

     Hence  0<t< pi/4 , 3*pi/4 <t <pi,  pi< t < 5* pi/4    and  7* pi/4 < t < 2pi (shown)
 
    (You may wish to draw the graph of v = 3cos (2t) to see this more clearly; the segments
    where the graph is above the horizontal axis are the relevant ones)   


(b) When the particle stops, 3cos (2t) = 0
    
    t = pi/4, 3*pi /4, 5*pi /4 , 7*pi/4  (shown)


(c) Area under velocity-time graph is equivalent to total distance travelled.
    
   Recognise that due to symmetry, we can simply find the area under the curve for
    
   v(t)= 3 cos(2t)  for 0<= t <= pi/4  and subsequently multiply this value by 4.


   Area under curve for 0<= t <= pi/4   =  int { from t=0 to t= pi/4 } 3 cos(2t)  dt

   = [ 3/2 sin(2t)] { upper limit: t=pi/4; lower limit: t=0)

   = 3/2

   Therefore, total distance travelled =3/2 *4  =6 units (shown)  

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