Math help plox

[kiljoy96]

I'm completely stumped on how to do this.
Basically, I'm in the very early research stages of a project. I've been doing physics and trig for the past 2 hours, and I just can't remember how to make an exponential graph, or even find out what I'm trying to find out.
At an acceleration rate of 229^2 inches a second, how much distance is covered between the dead stop and hitting 60 miles an hour? Is it just simply the acceleration rate multiplied by the time taken or something?

[redfield95]

Divide by zero :3

[i-am-ghey]

basic answer: consider the equation v^2-u^2=2as and plug in the numbers.

advanced answer: even though the space-time is assumed to be flat, not enough information is given. you did not state whether the velocity and accleartion is with respect to the inertia frame or not.

[tox]

At 6/17/11 05:31 AM, i-am-ghey wrote: basic answer: consider the equation v^2-u^2=2as and plug in the numbers.

advanced answer: even though the space-time is assumed to be flat, not enough information is given. you did not state whether the velocity and accleartion is with respect to the inertia frame or not.

Im a firing my ........ o.o......
....... ~ blinks a few times~
...... ~ walked out of the room and closes the door~

[MrPercie]

At 6/17/11 03:18 AM, kiljoy96 wrote: exponential graph

I probably do know how to do that but forgot, may wanna ask someone else.

At an acceleration rate of 229^2 inches a second

So basically do you increase your speed by 229 inches a second? so at 0 seconds your at 0 velocity, then at 1 second your at 229 inches per second, then 2 seconds later your 458 inches a second and so on until you get to 60 miles.

[MrPercie]

I worked it out to be 16600.87 seconds, anyone else?

[redfield95]

At 6/17/11 05:39 AM, MrPercie wrote: I worked it out to be 16600.87 seconds, anyone else?

I'm not good with math, so I can't help you

[MrPercie]

At 6/17/11 05:39 AM, MrPercie wrote: I worked it out to be 16600.87 seconds, anyone else?

crap, you said distance covered, not time taken.

let me think

[MrPercie]

At 6/17/11 05:42 AM, MrPercie wrote:

At 6/17/11 05:39 AM, MrPercie wrote: I worked it out to be 16600.87 seconds, anyone else?

crap, you said distance covered, not time taken.

let me think

fucking hell this is confusing. havent done maths in a while.

you may need some equatuions or something because im getting mixed up with acceleration, distance and velocity.

[i-am-ghey]

Lol thought I already answered the question

[MrPercie]

At 6/17/11 06:04 AM, i-am-ghey wrote: Lol thought I already answered the question

what does u, s and a mean? and what the hell are you talking about space for?

[GiantDouche]

At 6/17/11 06:11 AM, MrPercie wrote: what does u, s and a mean?

USA

[i-am-ghey]

U is initial velocity
v is final velocity which is zero
s is displacement

they are standard notations in physics you should know it

[MrPercie]

At 6/17/11 06:19 AM, i-am-ghey wrote:

they are standard notations in physics you should know it

well its not every bloody day I know the damn standard physics notiations

I did know V was velocity though.

[GiantDouche]

At 6/17/11 06:22 AM, MrPercie wrote: I did know V was velocity though.

LOOKS LIKE WE GOT OURSELVES A MOTHERFUCKIN' PHYSICIST

[VitaminP]

Also remember to convert your values into the same units i.e using miles and hours for all values or using inches and seconds for all values.

[Gimmick]

At 6/17/11 06:11 AM, MrPercie wrote: what does u, s and a mean?

USA

[Ninjafap]

Acceleration is constant. a(t) = a

The integral of acceleration is velocity (acceleration is the change in velocity). v(t) = a t

The integral of velocity is position (velocity is the change in position). p(t) = .5*a*t^2

Substitute velocity in for time. We get: p(t) = .5*a*(v^2/a^2) = .5*(229^2)*(63360^2/299^4) = 22452 inches. Convert it to 1.61 km which you should be using since this is physics.

If you want make a position versus time graph simply plug in your calculator y = .5*(229^2)*t^2 and stop it when time hits (63360^2/299^4) = .502 seconds.

Oh and take calculus. It's fun.

[Ninjafap]

What kind of trig were you trying to do and how do you know physics yet alone had this assigned if the position-velocity-acceleration relationship is one of the first things explained in any physics class (even if it's not justified with calculus)? Why did you type this out on Newgrounds instead of making a simple Google search?

[i-am-ghey]

um... accuracy might be a problem if calculations involve many steps.

it is unnecessary to take a long winded approach to solve this problem. one simple formula that any physics students are expected to know is sufficient.

of course, you can always consider partial derivatives to estimate the value if you don't have a calculator at hand. otherwise, i don't know why you should need to plot a graph at all.

[LostFaceInTrain]

Y'all gay here

[Ninjafap]

At 6/17/11 09:36 AM, i-am-ghey wrote: um... accuracy might be a problem if calculations involve many steps.

You're thinking of precision. And no, this is really just a one step plug-and-chug once you derive (or look up) the formula relating position to acceleration.

it is unnecessary to take a long winded approach to solve this problem. one simple formula that any physics students are expected to know is sufficient.

Exactly. I don't know if any one in this thread has taken physics, really.

of course, you can always consider partial derivatives to estimate the value if you don't have a calculator at hand.

There is only a single variable, time, so you can't take partial derivatives.

otherwise, i don't know why you should need to plot a graph at all.

I suppose it's to show how the position changes between the velocities of zero and sixty miles per hour.

[i-am-ghey]

there is a trick here.

s=u^2/2a, so in fact there are two variables.

you can take differential ds and consider small changes in u and a.

[Ninjafap]

At 6/17/11 09:48 AM, i-am-ghey wrote: there is a trick here.

s=u^2/2a, so in fact there are two variables.

you can take differential ds and consider small changes in u and a.

The final velocity isn't zero. Read his post again.

There aren't two variables because we know the acceleration. I suppose he's trying to make position the function of time, not velocity.

[i-am-ghey]

i supposed you haven't learnt this math tools yet. estimating the value of s is NOT about physics really.

i am a math and physics major.

also, i believe dead stop means 0 velocity.

[i-am-ghey]

ok. he means 0 to 60.

but the magnitude won't change by symmetry.

[JaY11]

Inches and miles per hour? Please tell me physics in the US is not based around the imperial system?

[Elfer]

Oh my fuck, you people need to stop dicking around and just answer his question.

KNOWN:
a = 229 in/s^2 (I'm assuming this is what you meant, because I don't know how acceleration in square inches works in reality)
v0 = 0 in/s
v1 = 60 m/h = 1056 in/s (ffffffffffFFFFFFFFUCK IMPERIAL UNITS)

UNKNOWN:
t1 (time to reach 60 mph)
D (distance covered between t0 = 0 and t1)

INITIAL CONDITIONS:
at t = 0, v = 0 and D = 0

DO SOME MATH:
a = 229 in/s^2
integrate w.r.t. time to find:
v = 229t in/s
and again:
D = (229/2)t^2 in
(both integration constants are zero due to the initial conditions, trivial calculations omitted)

Solve t1:
1056 in/s = 229(t1) in/s
t1 = 4.6114 s

solve D:
D = (229/2)(4.6114^2) = 2434.84 inches
= 202.9 feet
= 0.0384 miles

For fuck's sake

At 6/17/11 10:03 AM, JaY11 wrote: Inches and miles per hour? Please tell me physics in the US is not based around the imperial system?

Prepare to have your mind blown: Many engineers in the real world have to work around imperial units simply because a lot of instruments are designed or manufactured in the US.

[i-am-ghey]

so no one knows the magic formula v^2-u^=2as?

unbelievable.

[Elfer]

At 6/17/11 10:13 AM, i-am-ghey wrote: so no one knows the magic formula v^2-u^=2as?

unbelievable.

Why memorize formulas when you can derive everything with the power of CALCULUS

Also that formula is meaningless because you don't identify the variables. Convention is not the same everywhere!