Can the instantaneous velocity of an object ever be greater or less in magnitude than the average velocity?

Question

please explain or give an example?

Answer 1

Yes. Average velocity is determined from the slowest velocity, perhps starting from zero, and adding it to the greatest velocity, before it slowed down. So the average is the least and greatest added and divide by 2.

Answer 2

The instantaneous velocity of an object is its velocity at any specific point in time while an object is traveling between point A and B.

The average velocity is taking the speed at point A and then at point B and finding the average velocity for the entire time between A and B.

So yes the Instantaneous velocity can be both greater or lesser than the average velocity, assuming that the velocity changes from lesser than the average and greater than the average, in the time it takes to move from point A to B.

Example:
So If you have a car on a track, at point A, its instantaneous velocity would be 0 mph, and lets say it reaches a velocity of 100 mph by point B. So the average velocity assuming a linear rate of acceleration would be 50 mph.

Thus during the first half of the track, the Instantaneous Velocity would be Lesser than the Average Velocity, and near the last half it would be Greater.

Answer 3

Science Teacher needs to go back to school. The average velocity is derived from the equation v = d/t, where d = distance traveled and t = the time taken to get there; v, of course is the velocity in appropriate units.

So, if you travel halfway around a 2 mile track in 2 minutes your average velocity is 30 miles per hour (1mile/(1/30 of an hour)). If your goal is to average 60 mph you cannot do it, because you would have to travel the entire 2 miles in 2 minutes, and you have already used up that amount of time.

Hence, even if you were to go the rest of the way at 100 miles an hour you still average less than 60 mph. So, your instantaneous velocity in that instance (100 mph) will be much greater than your average velocity (less than 60 mph), and the velocity you maintained the first half of the journey will be less than the average.

Answer 4

Yes, consider an object that accelerates from rest with acceleration a for a time t, then deccelerates with acceleration -a for a time t until it come to rest. In the first time interval, it covers a distance:

d = 1/2*at^2 . By the symmetry of the probelm, it also covers the same distance during the decceleration. So teh total distance covered is 2d in a time 2t. Thus the average speed is:

= 2d/2t = d/t

Now the peak speed occurs at t before the object deccelerates. This speed is:

vp = at

Now unless a = d/t^2 the peak (Instantaneous) speed will be larger than the average speed.. Note that the instantaneous speed is also 0 at the start and end of the motion.

Answer 5

Let us consider four instantaneous velocities [v1, v2, v3 and v4] for an object which moves a distance of S in 4 seconds.

Distance moved in 1 second = v1 *1 = v1
Distance moved in the 2nd one second = v2 *1 = v2
Distance moved in the 3rd one second = v3 *1 = v3
Distance moved in the 4th one second = v4 *1 = v4


Distance moved for the entire four seconds = [v1 + v2 + v3 + v4] = S

The average speed [w] for the entire journey is

w = S / 4 = [v1 + v2 + v3 + v4] / 4. --------------------1

Now let us suppose that each v's in the R.H.S. are > w

We get

[v1 + v2 + v3 + v4] / 4. > 4w/4 > w

But from equation 1 this cannot be true,

Similarly all v's cannot be < w.

The possibilities are
1) all v's can be = w
or 2) some values must be < w and some values > w and some may be equal to w such that theirs whole sum divided by the number of instantaneous velocities must be equal to the average speed.

Instantaneous velocities of an object in a given interval of time or in a given distance may be greater or less or equal to the average speed for the entire path.

All of them can never be greater the average speed. Some may be greater.

All of them can never be less than the average speed. Some may be less.

Answer 6

Yes. Science teacher is otherwise wrong though. Average velocity is total distance divided by total time. (max+min)/2 is called midrange, not average. If something accelerates uniformly, then (max+min)/2 is average, but otherwise it's not.
For example, if velocity is 1 for the first two seconds and 3 for the next two seconds (I know that it cannot change instantaneously but I am trying to explain the concept of an average, not the physics of acceleration), then it travels 1*2+3*2=2+6=8 in four seconds, so the average is 2, but if it is 1 for the first one second and 3 for the next three seconds (again, I know that it cannot change instantaneously but I am trying to explain the concept of an average, not the physics of acceleration), then it goes 1*1+3*3=1+9=10 in four seconds for an average of 2.5.

Answer 7

The instantaneous velocity could be anything from 0 to light speed. If it's just for a very short period (like 1 nanosecond), it will not affect the average over (for example) 1 hour.

1 hour speed = 60 miles per hour.
instantaneous velocity at T1 = 10 minutes, 23.879443994 seconds = 99% of light speed.
instantaneous velocity at T2 = 55 minutes, 53.997361004 seconds = 0
Speed rest of time, except for a few nanoseconds around the times T1 and T2, speed = 60 miles per hour.

Answer 8

Yes, if it weren't the instananeous velocity would equal the average velocity throughout the process, and an object could never start from zero velocity and then attain zero velocity.

Answer 9

"average velocity" is displacement/time and "average speed" is distance/time. average velocity mentioned by Science Teacher is obtained only in a constant acceleration scenario. nonetheless, you can have an average velocity either above or bellow average speed.