why won't my (physics) equations work?

I am a creature of habit.
Once I have something memorized and have used it enough times to be comfortable with it, I tend to stick with it and not try to learn new ways of looking at things, because doing so tends to confuse me.
I have gone through almost an entire set of problems using the same set of two or three equations, getting all of them right, and then I get to the last 5 and they’re all wrong. I look at the solutions, and all of the ones I got wrong use an equation I’ve never seen before. But the problems seem just like the other 25 I’ve done with my other equations!!! ARGH!!! ripping hair
To shed some light on the situation:
The equations I’ve been using are these:

x (final) = x (initial) + [velocity (inital) * time] + [1/2(acceleration) * time^2]


velocity (final) = velocity (initial) + [acceleration * time].

The new equation that I’ve never seen before is this:

[velocity (final)]^2 = [velocity (initial)]^2 + [2 *acceleration * x (final)].

Where did this come from? Why would it work in a situation where the previous two equations wouldn’t? For that matter, in what situations would the previous two NOT work, and why?

Just so you know–this is not about getting this problem set done; it was due and was finished a long time ago; I am just genuinely perplexed by this and cannot go to sleep ignorant.

I’d appreciate any attempt at explaining this–if you would like the specific problems, just ask.

It works when you don’t know time.

Let me clarify, before I get mauled–
when I said I don’t try to learn new ways of looking at things, this ONLY applies to math & physics because they utterly dumbfound me. Not that I wouldn’t like to, it’s just that most of my attempts to do so end with me in the mental equivalent of a pitch black shitstorm.

when you don’t know time…
but why would the others not work?
I’ve had no trouble finding time or putting time in terms of other variables in other problems.

Are you talking about substituting with displacement/velocity ?

That works only if velocity refers to “avg. velocity” for the period and displacement the complete displacement ?

It looks like the third equation was generated from the other two to remove the time dependence. So the first one was solved for “time” then substituted into the other. x(initial) was apparently set to zero (since it doesn’t appear in the final equation). And I’m ashamed to say I actually went through the derivation just so I’d be sure. :frowning:

This seems more of a shortcut than anything else. Say you wanted to throw a ball straight up with some V[sub]initial[/sub] and wanted to know how fast it was going at a given altitude x[sub]final[/sub]. With just the first two equations, you’d have to plug all your known values into the first, solve for time, then plug that value into the second. That third equation has the “solve for time” built in, with the additional loss of the direction vector (technically, you get the speed).

As you say, there shouldn’t be any difference. Are you off by a sign (since the third equation only gives a scalar speed), or is it something more drastic?

no, I’m off by like an order of magnitude and then some. I checked my math, and it all looks good, but I’m going to keep trying it my way until I get the right answer if it takes me all night (morning…*sigh).

also, why does the third equation only give scalar speed? what about it makes it lose the direction vector?

(btw, thank you for the responses so far–I think they’re putting me on the right track.)

Wanna post an example problem and your solution?

ok…here’s one.

a hockey puck sliding on a frozen lake comes to rest after traveling 80 meters.
if its initial velocity is 3.29 m/s, what is its acceleration (in m/s^2) if that acceleration is assumed constant?

i used eq’n 1, plugging in x final as 80, x initial as 0, and v initial as 3.29. i used the 2nd eq’n to get t=3.29/a and plugged that in to eq’n 1, which eventually gave me an a of 0.20295, not the right answer even if I were to add a negative sign.
i can get the right answer using the 3rd eq’n, but to me that’s just something else to memorize, unless i can derive it, which i don’t know how to do.

(sorry i am not capitalizing properly–it bugs me but my boyfriend’s asleep about a foot & a half away from me so i’m trying to push as few buttons as possible)

Like the others have said…it takes the variable of “time” out of the equation…unless you’d rather solve two equations simultaneously to find “time”! Or, run through iteratations to find “time”! Talk about pulling your hair out! And, there will be no time just before/at an exam to start learning! You’ll be in a total panic.

You gotta be more willing to accept new methods of problem solving. There’s a good reason such equations are taught to us. And, if you are taking a college level course, profs love to not give you “time” just to see if they can trip you up.

When you are learning physics, always be alert. Remember - the profs (or teachers) have taught you everything YOU know…not everything THEY know! :wink: So, heed our words from experience!

Be alert! (…The world needs more lerts!)

  • Jinx

There are four such kinematic equations. Enjoy :wink:

You’ve just made an arithmetic error somewhere.

The answer using:

v[sup]2[/sup] = u[sup]2[/sup] + 2.a.s is

a = -0.06765 m.s[sup]-2[/sup]

If you do it your way, you get t = -3.29/a

80 = 3.29(-3.29)/a + 0.5.a(-3.29)[sup]2[/sup]/a[sup]2[/sup]
80 = -10.8241/a + 10.8241/(2.a)
80 = -5.41205/a
a = -0.06765 m.s[sup]-2[/sup]

Nitpick - You should correct that to

[velocity (final)]^2 = [velocity (initial)]^2 + [2 *acceleration * {x (final) - x(initial)}].

Also note that this is a scalar equation valid for straight line travel while the other one can be a vector equation.

There is missing a factor of 2, which is also Desmostylus observation.

I came up with:
[v=velocity; a=acceleration; t=time; s=distance; (t)=at time t; (0)=at time zero]

V(t) = v(0) + a*t

Now square this equation like (p+q)2 = p2 + q**2 + 2pq

v(t)2 = v(0)2 + (a2)*(t2) + 2*v(0)at

Now a little trick:

v(t)2 = v(0)2 + 2a( 1/2a(t2) + v(0)*t )

v(t)**2 = v(0)*2 + 2a( s(t) + s(t) )

v(t)**2 = v(0)**2 + 4a*s(t)
So, to me andy_fly is not correct, it is in both instances what he calls x(final).

Sorry, Mummy, but andy_fl is right.

Your equation
v(t)2 = v(0)2 + 2a( 1/2a(t2) + v(0)*t )
should become
v(t)**2 = v(0)*2 + 2a( s(t) - s(0) )

because s(t)-s(0) = 1/2a*(t**2) + v(0)*t

Hi Nevermore -

That third equation is derived from the first two that you gave.

v(f) = v(i)+at

Square both sides and you get v(f)^2 = v(i)^2 + 2v(i)at+(a^2)(t^2). Call this Equation 4.

Now take your first equation and set x(i) = 0 (you’re just calling the starting point “0”, an allowable simplification).

x(f) = v(i)t + 0.5a(t^2)

Double both sides and multiply by a and you get:

2ad(f) = 2v(i)at+(a^2)(t^2)

Substitute this into the last bit of Equation 4 and you get your third equation.

I solved your sample problem by saying that if we have constant decelleration from 3.29 m/s to 0 m/s then we can calculate time by dividing total distance over average velocity. v(a) = (3.29+0)/2.

80m/v(a) = 48.63 sec. That’s the time it takes the puck to stop while it travels 80m.

Knowing v(i) and t you can get the acceleration from v(i)/t.

1/t = 0.0205… could that be the number that you came up with for a the first time? Looks like just forgetting to include v(i).

FWIW, in high school physics, we had to memorize all 3 of those equations, and there were plenty of instances where one would work where the others would not. As others have said, the third one is for when you don’t know time.