physics nonsence?

Question

Can the average velocity of a particle be zero over a given time interval if it is not zero over a shorter time? Explain.

Answer 1

yes if you consider velocity a vector.

Answer 2

Absolutely. Average velocity is defined as:

V_avg = (final_displacement – initial_displacement) / Δt

where "final_displacement" and "initial_displacement" are vectors showing the particle's position at the end and beginning of the time period.

So this equation shows that, if the final displacement equals the initial displacement (that is, it the particle ends up in the same spot where it started), then the average velocity is zero. This is true regardless of how the particle may move in between the initial moment and the final moment.

Answer 3

Yes. Imagine you are a particle. You are standing on a field and walk due north at a speed of 2 mph and do this for 1 minute. Then you turn around and walk at 2 mph due south for 1 minute and end up right where you started. Your average velocity will be 0. Your average speed during that time will be 2 mph. Velocity is a vector measurement. You need to include the amount and the vector of the measurement. That is why you were walking 2 mph due south and not just 2 mph.

Answer 4

When a object is given power to move at time zero plus ,it is subject to maximum acceleration . As the mass moves it increases velocity non linearly in a particular direction.
What we measure between the two end points is really the average velocity . The reason is that we cannot measure instantaneous velocity.
The End point velocity of the moving object is called final velocity.
At the time zero plus the acceleration plots as maximum at that point the velocity is basically zero. then it increases as the acceleration decreases. This basically applies all to Inertial motion of all particle micromasses as well.
However in non inertial motion the Scenario is a little different;.
Where velocity increases with increased acceleration.
Therefore a micromass particle moving toward a gravitational mass will experience increased velocities due to the changing magnitures of the Gravity Field.

Answer 5

I would say that on a given time scale, a particles average velocity is zero if it's start position and end position are the same because velocity is supposed to be a vector measurement.

On a shorter time scale the average velocity could be different if the partical's end position at the shorter time scale is different than the start position and would then return to an average velocity of zero on the longer time scale once it returned to its original start position.

This, of course, all depends on your frame of reference.

Answer 6

Are the time frames' given? if not, then the answer is "of course". Average velocity could include periods of time when particle is accelerating as well as decelerating. Over the longer time period this could average to zero, by math alone. Over the shorter period (pick any one) the particle could be going faster or slowing down (not zero).

Answer 7

Velocity V is a vector, having both magnitude v and direction (i,j,k), where the i, j, k are so-called unit vectors that delineate direction in 3D space. [You can think of i,j,k as sort of shorter x,y,z.] Average velocity Vavg = (V +....+ V')/N; where N is the number of velocities in the average.

For example, if we average two velocity vectors, we can write Vavg = (V + V')/2 If the two velocites are relegated to just one dimension, say i, then Vavg = (V + V')/2 = (v i + v' i)/2; where v i and v' i means a speed of v and v' in the i direction.

Now, to your question...if Vavg = (v i + v' i)/2 <> 0 for delt < delT; where delt is a shorter elapsed time than delT, will Vavg = 0 for a longer period delT?

From Vavg = (v i + v' i)/2 we can see that Vavg depends on the magnitude of the velocities, v and v', and their directions (i). One way for Vavg = 0 is for v i = v' i; so that v i - v' i = 0. This can be written as v i + v' (-i) = 0 and the -i means a direction directly opposite to i with a plus sign. In this case Vavg = (v i - v' i)/2 = 0/2 = 0 for our example of two velocities.

So a necessary condition is that the two velocities be going in opposite directions. But that is not sufficient. From v i = v' i, we se also that v = v' because the i's cancel out. In other words, the magnitudes (speeds) of each velocity must be equal, too.

Notice that delT and delt are not a part of these equations for Vavg. That is, Vavg = 0 is independent of the elapsed times. This, of course, presumes the magnitudes and/or directions of the included velocities do not change.

So, no, if Vavg <> 0 over delt, then it will still be Vavg <> 0 over delT if the respective velocity magnitudes and directions do not change. But, yes, if Vavg <> 0 over delt, and the magnitude and/or direction of the included velocities do change, then Vavg could be = 0 over delT.

Answer 8

Yes if you factor in the direction of the object. The average can become zero, but if you look at a smaller time period, maybe in just one direction, then the average could be zero or something else. Again this depends on direction.

Answer 9

velocity is vector so its absolutely possible.

avg vel of a body in or circular motion SHM is zero because direction changes and the magnitude remains same so the vectors cancel out.

Answer 10

Maybe its speed varies? The averaging of positive and negative numbers can equal zero.