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a) A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 40 ft/s?
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 40 ft/s2. What is the distance covered before the car comes to a stop?
b) A car braked with constant dec. of 40 ft/s^2 and produced skid marks measuring 160 feet before coming to a stop. How fast was the car traveling when the brakes were first applied?
Okay, I'm not sure if I'm doing things right, but here goes my work.
The givens:
v = 50 mi/h
a = -40 ft/s^2 (because it is decelerating)
So, the first thing for me to do would be is to convert 50 mi/h into ft/s to get the correct units. So after dimensional analysis, I got: 73.33 ft/s.
Next, we know that dv/dt is the same thing as the acceleration. In our case, it is -40 ft/s^2.
So we use antiderivatives, and get v= -40t + c (this is after using antiderivatives). "c" here is a constant.
I do the same thing for the distance. dx/dt = -40t + c so I use antiderivatives again to get:
x= (-40t^2)/2 + c*t
Now, c, our constant can be found by substituting 0 in v=-40t+c. t = 0, velocity is 73.33, so c is 73.33
I'm CERTAIN that I use these formulas to solve the two problems. Correct me if I am wrong, because if I did this wrong, the whole problem is wrong.
So our two formulas are:
v = -40t + 73.33
and
x = -40t^2 / 2 + 73.33t
x is our distance while v is velocity.
I'm not sure what to do here because I thought for Part A, we solve for the time, then subsitute it in equation 2 to get the distance.
As for part B, we are given distance, so solve for time, and plug that value in Equation 1.
Is that right?
I think so, but I keep getting the wrong answer, I think they are wrong ><
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answer: i suck at SC... very much
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so in general
v = at + initial v
x = 1/2 * at^2 + vt + C (which in this case is x initial)
which can also be written as
delta x = vt + 1/2at^2
remember in your case a is negative
so im just gonna do b
if delta x is 160 ft and a is -40 ft/s^2
160 = -20t^2 + vinitial*t
vfinal = 0 = vinitial - 40*t
two equations, two unknowns, solve
and i dont have a calculator or mathematica on this comp so im not gonna do it
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yea the thing is that we cant use physics equations..we need to use antiderivatives :x
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On November 20 2008 17:29 clazziquai wrote:
yea the thing is that we cant use physics equations..we need to use antiderivatives :x
i did use antiderivatives, everything was derived from
v = vinitial + at
integral(v dt) = integral((vinitial + at) dt)
x = vinitial*t + 1/2at^2 + C
that's where it comes from in the first place dontcha know
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hm ok let me see what the f im doing wrong......:/
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Integration Integration Integration!
When... You... Integrate, acceleration, you get the change of speed
When you Integrate, velocity, you get the change of distance
Make sure you add initial speed
Make sure u add initial distance
Then you have it all in the bagggggg
Integration, a good friend to meee
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also raithed is a bitch
just had to say it
learn to not suck, faggot
EDIT: what a shitty fucking 3000th post dedicated to hwo much raithed sucks
oh well
it means he really really really sucks
assface cuntbag bitch go kill yourself if you're not already braindead, i can't really tell
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ok SO my equations were right, now imma try this again.
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Tuna, welcome to the Terran Ghost icon club!
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On November 20 2008 17:37 clazziquai wrote:
Tuna, welcome to the Terran Ghost icon club!
bitch im switching races
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LOL thanks dude.
I did have the equations, the biggest mistake I forgot to do was to set the final velocity equal to 0.
FFS IM I AN IDIOT
FUCK ME. Sorry for wasting your times goddamn -_-
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it's cool as long as you're not raithed
fuck that fucker seriously
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Oh wow I've never heard of antiderivatives before, and I'm doing a degree in Maths.
But it appears to just be an integral?
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On November 20 2008 23:33 Nytefish wrote:
Oh wow I've never heard of antiderivatives before, and I'm doing a degree in Maths.
But it appears to just be an integral?
Yes, it is. And Clazziquai, you need to find a smart person in your class and make friends with that person.
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On November 21 2008 00:09 Chromyne wrote:Show nested quote +On November 20 2008 23:33 Nytefish wrote:
Oh wow I've never heard of antiderivatives before, and I'm doing a degree in Maths.
But it appears to just be an integral? Yes, it is. And Clazziquai, you need to find a smart person in your class and make friends with that person.
haha why so??
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United States17042 Posts
I think that you've got a handle on it, don't worry about it clazziquai.
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On November 21 2008 00:46 waterGHOSTCLAWdragon wrote:
I think that you've got a handle on it, don't worry about it clazziquai.
I love you Ghostclaws
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On November 21 2008 00:11 clazziquai wrote:Show nested quote +On November 21 2008 00:09 Chromyne wrote:On November 20 2008 23:33 Nytefish wrote:
Oh wow I've never heard of antiderivatives before, and I'm doing a degree in Maths.
But it appears to just be an integral? Yes, it is. And Clazziquai, you need to find a smart person in your class and make friends with that person. haha why so??
Fundamental theorem of calculus, anyone?
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EPT![[image loading]](http://img260.imageshack.us/img260/9026/stfuvv0.gif)
