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/sci/ - Science & Math

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1106009 No.1106009 [Reply] [Original]

A car traveling at 41 ft/sec decelerates at a constant 5 feet per second squared. How many feet does the car travel before coming to a complete stop?

>> No.1106018

>ft
>homework
double gtfo

>> No.1106034

Around 3 feet I'd say

>> No.1106056

41s - 5s2 = 0

>> No.1106060

The equation you're looking for is
<div class="math">v_2^2 - v_1^2 = 2a(s_2 - s_1)</div>

>> No.1106075

>>1106056
what do I do with that then

>> No.1106093

336.2ft

>> No.1106110

yep not working sorry. any other ideas

>> No.1106117

1) SI units dickwad.
2)SUVAT.

>> No.1106121

>>1106075
I have no idea I just wrote it out as a Algebra 1 problem.

>> No.1106123

SI UNITS MOTHERFUCKER, DO YOU USE THEM?

>> No.1106138

>5 feet per second squared
25 feet per second

>> No.1106144
File: 17 KB, 527x434, 1275585529387.png [View same] [iqdb] [saucenao] [google]
1106144

>>1106060
>>1106139

>> No.1106139 [DELETED] 

on average he'll be going <span class="math">(v_{avg} = v_{max}/2)

The time to stop is <span class="math">(T = v_{max}/a)

So the distance traveled is (X = v_{avg} T = v_{max}^2/2a)[/spoiler][/spoiler]

>> No.1106158

on average he'll be going <span class="math">v_{avg} = v_{max}/2[/spoiler]

The time to stop is <span class="math">T = v_{max}/a [/spoiler]

So the distance traveled is <span class="math">X = v_{avg} T = v_{max}^2/2a[/spoiler]

>> No.1106159

From >>1106060
<div class="math">0^2 - 41^2 = 2 \cdot -5 \cdot (s_2 - 0)</div>

You should be able to figure it out now. -_-

>> No.1106174

>>1106093
Whoops, my bad, it's half of that, lol. 168.1 ft

>> No.1106231

>>1106158
This is real physics. Thinking a problem out logically. Fuck your equation plugging.

>> No.1106256

don't you learn this in like the first week of any intro/ high school physics class?
also
>itt: implying OP chose what for his homework to use feet

>> No.1106259

>>1106075
v2=speed after deceleration (the car stops in the end, so 0)
v1=speed before deceleration (41 ft/sec)
a=acceleration (since it's decelerating, it's negative: -5 ft/sec^2)
s2=distance at which the car stops (our x)
s1=distance at which the car starts decelerating (0, since you want to know the distance from where the car starts decelerating)
So, s2=(v2^2)/(2a) = -(41 ft/sec)^2/2x(-5 ft/sec^2) = -(1681 ft^2/sec^2)/(-10 ft/sec^2) = 168,1 ft

>> No.1106266

>>1106231
Presuming you meant >>1106159

I thought it out, I just didn't explain the thought process that selected that particular formula as being the most useful. The formula itself should be memorized and chilling in the back of one's head for use (naturally, you could memorize the differential <span class="math">v dv = a ds[/spoiler], but in the case of a constant acceleration and inertial reference frame, this simplifies to the above).

Re-deriving the equations is silly. Why reinvent the wheel, anon?

>> No.1106280 [DELETED] 

>>1106266
WTF? Why don't you look at the post I was actually referring to?

>> No.1106310

>>1106266
You are mindlessly sticking in numbers into an equation, even the zeros. That is not physics, it's even worse than engineering and accounting. That's how you learn the subject - by reinventing every solution to every problem in different and unique ways. Next you can solve it by actually integrating. Each gives you a different perspective and intuition about the physics involved. Solving a general form and plugging shit in later is the bottom of the barrel.

>> No.1106338

>english instead of metric
Get out of here americunt.

Hey, how much is a mile?
>dur a lot

>> No.1106345

>>1106338
I'm not going to learn the metric system just for the sake of my AP physics class retard.

>> No.1106346

>>1106310
I "mindlessly" stick them in because that's good accounting practice. Sometimes you don't have all the variables required and so you need to introduce a constraint or an assumption into the problem. The fact that the 0s are even in the equations shows that I understand that when a car stops, the velocity is 0; additionally, I must also set my reference frame such that either s1 or s2 is 0, else the equation is impossible to solve (or I forgot a constraint).

Mindlessly? You're pathetic.

_If_ I had needed to, I might have been able to get the answer from a solution such as >>1106158 , but that's time and effort in my case that a) isn't worth spending time on and b) is in a general case incorrect (which is the reason why I included vdv=ads), so why bother working from _that_ as a solution?

Real physics isn't any of this. Real physics is taking a particle/body measuring these values so we can model the deceleration/velocity/whatnot and then glean useful information from that. This... this is high school physics.

>> No.1106349

>>1106346 samefag
and measuring*

>> No.1106353

*sigh*

Average speed is (41 + 0)/2 = 20.5.
Time it takes for car to come to a complete stop = 41/5 = 8.2.
20.5 * 8.2 = 168.1.

sources:
my infallible common sense

>> No.1106359

>>1106345
And you are an idiot. The world today increasingly uses metric as its standard. To not learn it is to handicap yourself when you're out of school.

Good luck.

>> No.1106369

>>1106346
If this is a subset of a larger problem, I'll use the equation and move on. If this is the problem in itself, it's embarrassing to "solve" it by using that equation and you should be embarrassed.

>> No.1106373

>>1106359
I was trying to laughably imitate OP. I thought that was obvious.

>> No.1106408

>>1106346
>>1106369
I suppose I should add my reasoning for that particular formula: I have velocities, I have an acceleration, and I have a distance. The problem asks for a distance.

Why introduce time as a factor into my equations? I could necessarily derive it if I so chose (as has been demonstrated), but this is going the long way about the problem.

It is embarrassing to be inefficient.

My job, were I the OP, would be to learn it by myself or to ask his teacher and classmates rather than posting to a board known for its generally lower IQ population compared to his teacher or his fellow classmates.

My job is not to be the OP. My job is to solve the equation and move along, or not post at all as the rules would see fit. Do I need to teach the OP? No, because that would take more time then I care to spend on an Internet board, and I've already done that for students at my school (and actually love doing so, as I love seeing the light bulb in another person). I tutor at my school; let me assure you that I would not have simply given the answer away without teasing out the necessary connections from the student I was aiding.

You assume too much. You are wrong to do so.

>>1106373
Implications are so useful in this regard. -_-

>> No.1106462

>>1106408

Why introduce time? Because time exists and plays a role in nature. It adds an additional perspective to the problem. Again, a raw kinematic equation is one method, another is the intuitive approach.

I also tutored math and physics in my community college, and I also lectured & led the lab section for undergraduate E&M at Berkeley as a grad student. This is exactly why sticking numbers in an equation just doesn't cut it. They damn well groked the capacitance of a parallel plate capacitor since I derived it a handful of times in different ways. I'm glad you don't give that away.

>> No.1106488

>>1106462
Introductory E&M, by the way. Non-calculus, but they knew enough to do the integrals.

>> No.1106517

v=u+at
t=(v-u)/a
s=ut+(1/2)at^2
s=u(v-u)/a+(1/2)a((v-u)/a)^2
v=0
s=u2/a-(1/2)u2/a
s=(1/2)u2/a=(1/2)*41*41/5=168.1ft