Calculating distance

muhammad26 said:
Calculate the distance traveled over the given intergral.
v(t) 4cos 2t, 0<t<2pi
Click to expand...
What are your thoughts regarding the assignment?

Please share your work with us ...even if you know it is wrong.

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muhammad26 said:
Calculate the distance traveled over the given intergral.
v(t) 4cos 2t, 0<t<2pi
Click to expand...

I suppose the question is, "Calculate the distance traveled over the given time interval, given that v(t) = 4 cos(2t) for t from 0 to 2pi"

The basic thing you need to be aware of is that the distance traveled is not the same as the displacement (net change of position). The object will have moved both forward and back during this time, and you need to calculate the (absolute value of the) distance traveled in each direction, and add them up. Does that sound like something you have been taught?

Give it a try and show us your work, so we can see where you need help.
 
muhammad26 said:
Calculate the distance traveled over the given intergral.
v(t) 4cos 2t, 0<t<2pi
Click to expand...
Note that the object can only change direction when velocity = 0. It does not have to, though. A car stops at many stop signs and then continues in the same direction after they stop. BUT when a car does change directions it must stop first.

So if this objects starts at point A and finishes at point B the distance traveled may NOT be |B-A| (the distance between B and A. Why is that? Let's suppose you were on a number line and went go from 0 to 3 then stopped and went to 1, then you stopped and then went to 7, then you stopped again and went to -1, stopped and then ended at 2. The distance traveled is not from just 0 to 2 which is 2.

The distanced traveled is rather |3-0| + |1-3| + |7-1| + |-1-7| + |2- -1| = 3 + 2 + 6 + 8 = 19


So you need to find out where v = 0 and partition your interval, which is 0 to 2pi, wherever you stopped. So step one is to find where v = 0, step 2 is to partition the interval based on where v=0 and step 3 is to find the distance traveled over each distance (distance is never negative) and finally step 4 is to add up all these distances
 
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