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Translational Motion Equation Derivation

by helpmeplzzz
Tags: derivation, equation, motion, translational
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helpmeplzzz
#1
Jul6-12, 08:38 PM
P: 8
1. The problem statement, all variables and given/known data

I'm studying for my MCAT and currently on translational motion. Rather than memorize the formulas, which would be useless for this type of test, I'm trying to learn and understand where the formulas come from, which requires knowing how to derive them. I need help with deriving the following translational motion equation:

2. Relevant equations

vx2 = vo2 + 2axΔx


3. The attempt at a solution

This equation is used for solving Physics problems in which time is not relevant. I tried substituting equivalent expressions for time (e.g. Δv/a and Δx/v) into other translational motion equations, but it was no use. =( I'm wondering if it's a Calculus based derivation? Or perhaps I am over thinking this and it's a common sense formula that has no derivation?

Please help!
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CFede
#2
Jul6-12, 09:03 PM
P: 42
It's actually a bit of both things you did. Start with this equation

[itex]x(t)=x_0+v_0 t+\frac{1}{2}at^2[/itex]

Then replace [itex]t=\frac{\Delta v}{a}[/itex], (wich is obvious from the definition of acceleration) so you get

[itex]\Delta x = x(t)-x_0=\frac{v_0\Delta v}{a}+\frac{\Delta v^2}{2a}/[itex]

Now, just remember that [itex]\Delta v=v_f-v_0[/itex] and multiply by 2a:

[itex]2a\Delta x=2v_0v_f-2v_0^2+v_f^2-2v_0v_f+v_0^2[/itex]

[itex]\Rightarrow v_f^2=v_0^2+2a\Delta x[/itex]

and ther eyou have it :D

Hope it helps.
helpmeplzzz
#3
Jul6-12, 09:27 PM
P: 8
YES!!! Thanks so much! That's exactly what I did before, but I didn't FOIL the Δv2, I just had vf2-vo2.

PhanthomJay
#4
Jul6-12, 10:11 PM
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Translational Motion Equation Derivation

It is wise to understand where the kinematic equations for constant acceleration come from. Once you understand them, it is essential to memorize them. It is not a useless exercise to do so.
azizlwl
#5
Jul6-12, 10:57 PM
P: 963
[itex]s=\int_{Vi}^{Vf} \! v \, \mathrm{d} t[/itex]
dv/a=dt
a is constant.

[itex]as=\int_{Vi}^{Vf} \! v \, \mathrm{d} v[/itex]
2as=Vf2-Vi2
Muphrid
#6
Jul7-12, 12:26 AM
P: 834
All that's needed here is conservation of energy. Starting with a constant force (and hence, acceleration), conservation of energy tells us that initial kinetic energy + work done on the particle = final kinetic energy, or

[tex]E_0 + W = E_f \implies \frac{1}{2} m v_0^2 + m a \Delta x= \frac{1}{2} m v_f^2[/tex]

Cancelling the mass and moving terms around easily gets you the correct result.

Most of basic kinematics comes from conservation of momentum ([itex]F = dp/dt[/itex]) or conservation of energy, really, and what's taught just corresponds to simple cases of what form the force might take. For instance:

1) No force (F=0): particles move in straight lines
2) Constant force: particles move along parabolas
3) Force proportional to position: springs
4) Central forces: Newtonian gravity, circular motion

In the end, whatever the kind of force being studied or what mathematical form it takes, someone somewhere solved the differential equation [itex]F = dp/dt[/itex] for each general kind of force of interest. When combined with the principle of conservation of energy, a scientist has a very powerful set of ideas at their disposal to analyze physical problems. I don't think everyone needs to be able to derive the equations of motion for any arbitrary force, but knowing how to go from these principles to the simple case solutions is useful, and if you face a system with a force like nothing you've seen before, sometimes the only thing to do is to fall back on these basic laws


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