Why does zero average velocity equal zero displacement?by opticaltempest Tags: average, displacement, equal, velocity 

#1
Jan1106, 06:14 PM

P: 135

I have a homework question which I would like to try to explain more mathematically.
If the average velocity of an object is zero in some time interval, what can we say about the displacement of the object for that interval? Thinking about it graphically, I realize it must be zero. Could anyone give me a hint as to how I could explain this using algebra and calculus? 



#2
Jan1106, 06:20 PM

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[2+(2)]/2 = the average = 0
if an average of velocities ends is ends up being zero it must mean some of those velocities must be negative. What does a negative velocity mean? try first explaining this mathmatically for two velocities such as 2 and 2 that are each constant during half of the trip. Find the average velocity from the two velocities that the object travelled at. 



#3
Jan1106, 06:22 PM

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Zero average velocity does not equal zero dislacement.
Zero average velocity equals zero average displacement. :) 



#4
Jan1106, 06:26 PM

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Why does zero average velocity equal zero displacement? 



#5
Jan1106, 07:08 PM

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Start with the definition of average velocity [where the timeweighted aspect is often forgotten]:
[tex]v_{av \mbox{\scriptsize\ from A to B}}=\displaystyle\frac{\int^{t_B}_{t_A} v\ dt}{\int^{t_B}_{t_A} dt}[/tex] Can you interpret the numerator and denomenator? If I recall correctly, the "displacement for that interval" depends only on the endpoints of the interval in question. 



#6
Jan1206, 03:29 PM

P: 343

No Tide, Dear understand that displacement is something that depends only on the final and initial state. The nonzero part is the distance covered which is a scalar and hence always positive. There is no such thing as average displacement. I don't what you to have such conceptual defects.
Robphy has done it anyway. Average velocity = displacement/time. Average speed = distance/time and even here if average speed is 0, distance covered is also 0. 



#7
Jan1206, 11:48 PM

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[tex]<\vec D> = \frac {\int_0^t \vec D(t') dt'}{\int_0^t dt'}[/tex] 



#8
Jan1306, 02:07 PM

P: 343

Don't argue on this dear, Displacement = final position  initial position.
I don't find anything differant in the integral equation you gave. The answer only depends on that 0 and t  more precisely the interval through which you are integrating. 



#9
Jan1306, 02:34 PM

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#10
Jan1306, 11:03 PM

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Average displacement is most certainly a valid quantity. If, at n different moments, the displacement is [itex]\vec D_j[/itex] then the average displacement is [tex]\frac {\sum_{j} \vec D_j}{n}[/tex] 



#11
Jan1406, 03:37 PM

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Let's say I take my mouse from its center position (from which we'll define the displacement) and move it quickly to the right some arbitrary distance and then equally quickly back to the center. Then, I move it slowly the same distance to the left and equally slowly back to center again. Over this entire time, since the motions are symmetric (that is, the mouse spends the same amount of time moving in the positive direction as the negative), the average velocity must be zero. Also, we're back to the central position, so the net displacement must be zero. However, since we spent more time moving the mouse on the left, the timeaveraged displacement must point to the left. Would you agree? 



#12
Jan1406, 04:55 PM

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ST,
Yes, good example! In fact, you don't even need the second part to make the point. The excursion to the right and then back to the "origin" is sufficient to demonstrate zero average velocity but nonzero average displacement since the displacement is nonnegative during that interval. I withdraw my illconsidered assertion and thank you for setting me straight. :) 


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