1-D Kinematics

It'd help if I knew exactly what you were trying to do.

If you solve and get two times, both are correct answers unless one is outside the range of applicability of the problem.

In just about any case, stating d = ut + a(t^2)/2 is technically wrong. It'll only apply between an initial and a final time (say between when an object is thrown/launched/released/whatever and when it lands or hits something). Before the initial time the equation for d is different (a constant if it's a rock in your hand for example). YOU have to determine which solutions, if any, are out of context. Who told you to take the positive "V2" anyway?
 
kingwinner,

I don't mean to be alarmist, but you are doing all kinds of illegal mathematical manipulations with your vectors. You cannot, for example, divide the vector v2 - v1 by the vector a to get the scalar Δ t. You also cannot put those vectors inside the square root as you have done. Who told you this was ok? It is true that v2 - v1 = a Δ t, but it isn't meaningful to divide by a vector quantity. This is just the definition of acceleration in terms of velocity i.e. acceleration is the rate of change of velocity. For uniform acceleration the rate of change of velocity is constant, and hence the velocity is a linear function of time.
 
Hi Physics Monkey,
Unless I miss my guess, kingwinner is dealing solely with the magnitudes of the velocity vectors. I think that eir class hasn't yet dealt with vectors properly; that ey might be currently learning them in maths but have not yet used them explicitly in physics.
 
Hi Pete,

I feel like you're probably right, I was simply alarmed when I saw all those vector symbols being manipulated so casually. I know from my own experience teaching students vector calculus just how hard it is to get various bad habits of vector manipulation unlearned.
 
Well its in the title of the thread: "1-D Kinematics"

The equations are all correct as far as I can see, except maybe for the arrows above the variables.

kingwinner: I just realized what you were doing - I previously didn't see that V2 was a substitution simply applying another kinematics equation.

The equation doesn't give +/- V2, it gives two solutions for V2 (though its true that one will be a positive number and the other a negative number). Remember you substituted "V2^2" for "V1^2 + 2 a d," so to apply your equation for "a" you need to calculate V2, which depends on "a" because of the substitution. One of the two V2s will give you back the acceleration you started with, the other won't.

I hate square roots...
 
przyk said:
Well its in the title of the thread: "1-D Kinematics"

The equations are all correct as far as I can see, except maybe for the arrows above the variables.

kingwinner: I just realized what you were doing - I previously didn't see that V2 was a substitution simply applying another kinematics equation.

The equation doesn't give +/- V2, it gives two solutions for V2 (though its true that one will be a positive number and the other a negative number). Remember you substituted "V2^2" for "V1^2 + 2 a d," so to apply your equation for "a" you need to calculate V2, which depends on "a" because of the substitution. One of the two V2s will give you back the acceleration you started with, the other won't.

I hate square roots...
Click to expand...

Thanks for explaning, but I don't quite get it :confused: .....whywould there be two solutions for v2? and why is the -v2 discarded?
 
When you square V1, as you do in V2^2 = V1^2 + 2ad, you lose information about V1, so its not surprising that you get two solutions for V2: one for each of the two V1s that you could have squared to get V1^2.

You - are - making - me - THINK.

STOP IT!!!!!!!!!!!!!!
 
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