Flatland
Hi everyone!

[Before to ask for help I would like to say "I don't speak English very well" because English is not my mother tongue , so if I make a mistake writing I'm really sorry ]

I have a question about Average velocity formula because there are two formulas and I don't know which one use for to solve problems and why there are two, The formulas are:

Average Velocity= V1+V2/ 2
Average Velocity= X-X0/T-T0

Why there are two formulas? What is the difference between the first one and the second one?

Thank you very much !

Have a nice day!
Robittybob1
Do they give a different result? There could be two formulas, since "Velocity" is a calculated quantity. Velocity is not something that can be directly measured.
rpenner
There are two different formulas because there are two different states of knowing things.

And you have made a huge error in not writing down the parentheses or the conditions.

For constant acceleration, A, we have

X = X0 + V0 (T − T0) + (1/2) A(T − T0)˛
V = V0 + A (T − T0)

So
X1 = X0 + V0 (T1 − T0) + (1/2) A(T1 − T0)˛
V1 = V0 + A (T1 − T0)

So
(V0 + V1)/2 = ( 2 V0 + A (T1 − T0) )/2 = V0 + (1/2) A(T1 − T0) = (X1 - X0)/(T1 − T0)

So when acceleration is constant, the average of the instantaneous velocity at times T0 and T1 is (V0 + V1)/2, and this is equal to the net distance traveled (X - X0) divided by the elapsed time (T − T0).

Thus the average velocity that the object moved at is the average of the instantaneous velocities at the end points -- as a consequence of the acceleration being constant.

As for Robittybob1's claim that velocity is not something that can be directly measured, that is not a well-supported position and has nothing to do with the physics that you are trying to learn.
Robittybob1
QUOTE (rpenner+Jun 30 2012, 05:35 PM)
....
As for Robittybob1's claim that velocity is not something that can be directly measured, that is not a well-supported position and has nothing to do with the physics that you are trying to learn.

What I was thinking is to measure velocity don't you measure distance traveled in a certain time, so you need to have two measurements before you do the calculation. So is that a direct measurement?
rpenner
In a mirror's co-moving frame of reference the magnitude of momentum in equals momentum out. So if the momentum of light sent to a mirror doesn't match the momentum of returning light that tells you something about relative motion of the mirror.

This is true with even individual photons.

So instantaneous velocity seems to strongly represent a measurable property of objects in the real universe.
Robittybob1
So when a mirror is moving toward the beam of light is the momentum in higher than momentum out?

I would be very interested to see a link to something I could read up on this please.
boit
Momentum must be conserved and ya all know that all ballistic theories of light ware disapproved long time ago. A moving mirror can only impart momentum to the photon, it can't make the photon move any faster. A photon is massless and things that are massless at rest can never be accelerated. They have momentum though whose magnitude can increase in accelerated frames. It will register as higher frequency. A Doppler shift of sorts.
rpenner
Photon momentum is proportional to their energy which is proportional to their frequency -- but bringing up that photons with different momenta all travel at the same speed is getting to be off-topic for a thread started by a first-week question on Newtonian kinematics.

If you bounce a ball off of a stationary wall, it bounces back with (approximately) the same horizontal momentum. If the wall is moving toward you then the ball bounces back harder than you threw it. While very few actual walls move, the principle is why home runs travel further than fastballs.
Robittybob1
QUOTE (rpenner+Jul 4 2012, 06:24 AM)
Photon momentum is proportional to their energy which is proportional to their frequency -- but bringing up that photons with different momenta all travel at the same speed is getting to be off-topic for a thread started by a first-week question on Newtonian kinematics.

If you bounce a ball off of a stationary wall, it bounces back with (approximately) the same horizontal momentum. If the wall is moving toward you then the ball bounces back harder than you threw it. While very few actual walls move, the principle is why home runs travel further than fastballs.

With the ball the speed is faster after hitting the moving wall. You could say the closing speed of the wall - ball is faster when the wall is moving toward the ball.

Faster after, not before hitting the wall. OK.

But if we replace a ball with a photon and the wall with a moving mirror the frequency of the light is higher coming off the mirror compared to the incoming frequency.

Yet is true that if the mirror was an observer the frequency of the light would be higher even before it hit the mirror? (Terrell Rotation effect of observing incoming light being blue-shifted)
boit
Funny thing is that individual photon will degrade. I saw in another site someone (not an authority) stating that if anything the photon will red shift after bouncing off a mirror (the mirror was stationary in their case).
Simple Netwnonian Mechanics: ball moves faster after colliding with a moving wall
Elementary Einstein relativity: photon shift frequency after bouncing off a moving mirror (a la speed gun radio waves).
rpenner
That makes no sense. If the source of the claim is not an authority, why do you take the source's statement as factual?
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