# Velocity change by force in one and two inertial frame confusion

JordanGo
Hi,
I just finished class and my professor was writing some of Newton's Laws on the board and derived some equations. We ended up with:
V(Δt)=FΔt (this is for velocity in first inertial frame
V(2Δt)=2FΔt (this is for velocity in second inertial frame​
Then he went and got the position in respect with time:
x(Δt)=F(Δt)(Δt) (which is just the integral of velocity in first i.f.)​
But then he did it for the second inertial frame and got:
x(2Δt)=x(Δt)+2F(Δt)(Δt)​

I do not understand where the x(Δt) comes from... Can anyone clarify this for me?

Homework Helper
Don't understand your notation ... do you intend $V(\Delta t)$ to mean that $$V$$ is a function of $$\Delta t$$ ?

Perhaps if the context were provided?

JordanGo
Yes, the velocity is a function of time.
When a constant force is applied, the velocity changes with respect to time.

Homework Helper
So what could $$F(\Delta t)(\Delta t)$$ ...mean? Or $$F(2\Delta t)$$ ??

JordanGo
The velocity is equal to the force multiplied by the time. When you take the integral of the velocity with respect to time, you get the force times the time squared. Now, for the second inertial frame velocity, I do not understand where he gets the equation (V(2t)). I think he means to generalize it but I'm not sure.

Homework Helper
If we write v=at in the first frame, the surely $x=\frac{1}{2}at^2$?

What is the relationship between the frames?

This is murder on a tablet... I'll go get a lappy.

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Homework Helper
Lets see... it looks like your professor is ignoring constants on order to focus your attention on the relationships.

Usually, two reference frames are moving with some relative speed.

If the speed of an object in frame 1 is $v_1 = Ft/m$ (because F=ma) then he could be saying that the speed in frame 2 is $v_2(t)=v_1(2t)$ i.e. the same as in frame 1 but with the time axis stretched out.[*] notice how this means that the initial speed of the object is zero in both frames? Is this possible?

... this would mean that $v_2=2Ft/m$ so, the frame 2 observer reckons the force is twice what the frame 1 observer says.

It also means the speed between the frames is $u=v_2-v_1$ ...

In frame 1, $x_1(t) = \frac{1}{2m}Ft^2 + x_1(0)$ as normal - your professor is leaving off all the subscripts and constants.

For frame 2, you have the same derivation - I think that the secret to understanding what he's written is to go through it again yourself, being careful about the constants and the notation.

This whole thing has been due to defining all terms in frame 1 and then using them in frame 2.

Not, on the whole, the approach I would have chosen - that goes double if the subject is special relativity ...

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[*] I will try to only use brackets to indicate functions ... part of the confusion came from the use of two different notations in different parts of the same equation.

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