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ljstella
09-13-2009, 05:33 PM
The driver of a car going 88.0 km/h suddenly sees the lights of a barrier 38.0 m ahead. It takes the driver 0.75 s to apply the brakes, and the average acceleration during braking is -10.0 m/s2.

He will hit the wall.

What is the maximum speed at which the car could be moving and not hit the barrier 38.0 m ahead? Assume that the acceleration doesn't change.

//<86TA>\\
09-13-2009, 05:50 PM
threve

ljstella
09-13-2009, 05:58 PM
threve

...what?

Tsar
09-13-2009, 06:01 PM
57.

ljstella
09-13-2009, 06:04 PM
57.

No...

Tsar
09-13-2009, 06:09 PM
Yes

ljstella
09-13-2009, 06:13 PM
Yes

The website rejected the answer. So, I guess not.

I hate WebAssign :(

FlyingDutchman
09-13-2009, 06:14 PM
wait does he hit the brakes with 38.0m to go or is 38.0m then he hits the brakes?

Tru2Chevy
09-13-2009, 06:15 PM
wait does he hit the brakes with 38.0m to go or is 38.0m then he hits the brakes?

When he sees the barrier he is 38m away, then .75 seconds later he actually applies the brakes.

- Justin

Tsar
09-13-2009, 06:16 PM
Ok try 58.

FlyingDutchman
09-13-2009, 06:17 PM
When he sees the barrier he is 38m away, then .75 seconds later he actually applies the brakes.

- Justin

ok thought soo

ljstella
09-13-2009, 06:19 PM
Ok try 58.

Not that either... :(

Tsar
09-13-2009, 06:19 PM
59?

ljstella
09-13-2009, 06:21 PM
59?

...are you just guessing?

bubba428
09-13-2009, 06:22 PM
idk tsar...I came up with 59.4km/h

Tsar
09-13-2009, 06:23 PM
...are you just guessing?
Well the answers are being beamed down to me from the Space Station via the satellite, so the numbers come in a little bit distorted.

Just trying to help.

FlyingDutchman
09-13-2009, 06:24 PM
w00t

bubba428
09-13-2009, 06:24 PM
at least it didn't come in as 75 :lol:

ljstella
09-13-2009, 06:25 PM
Well the answers are being beamed down to me from the Space Station via the satellite, so the numbers come in a little bit distorted.

Just trying to help.

No its okay, its just that I have a limited number of tries...

Tsar
09-13-2009, 06:27 PM
How many more do you have left?

FlyingDutchman
09-13-2009, 06:28 PM
How many more do you have left?

59...

Tsar
09-13-2009, 06:34 PM
59...
No, you lie.

FlyingDutchman
09-13-2009, 06:36 PM
wait i think i got it... the maximum speed he can be going and stop in time is 19.84 m/s or 71.4 km/h

try that i misunderstood the first time lol

using

d = (v)(t) + (1/2)(a)(t)^2


thats how i came up with my number . . .

v = 88 km/h = 24.4 m/s reaction = 0.75s distance before brakes = d = (24.4 m/s)(0.75)s d = 18.3m to hit brakes total 38.0m - 18.3m leaves 19.7m to stop

t = sqrt (d)(2)/(a) t = sqrt (19.7)(2)/(10) t = 1.98s to stop in time

maximum Velocity he can go using d = (v)(t) + (1/2)(a)(t)^2

19.7 = (v)(1.98s)+(0.5)(-10m/s^2)(1.98s)^2
19.7 = 1.98v - 19.602 39.302 = 1.98v v = 19.84 m/s

Tsar
09-13-2009, 06:39 PM
wait i think i got it... the maximum speed he can be going and stop in time is 19.84 m/s

try that i misunderstood the first time lol

using

d = (v)(t) + (1/2)(a)(t)^2
That's not correct.

FlyingDutchman
09-13-2009, 06:47 PM
That's not correct.

how so?

ljstella
09-13-2009, 06:48 PM
Well, I just got locked out, just before I tried something I came up with... I just emailed the teacher to see if I can get another few tries... ugh :( thanks anyway

FlyingDutchman
09-13-2009, 06:49 PM
Well, I just got locked out, just before I tried something I came up with... I just emailed the teacher to see if I can get another few tries... ugh :( thanks anyway

whatd you try?

ljstella
09-13-2009, 06:58 PM
whatd you try?

I tried using Velocity Final squared is equal to Velocity Inital Squared plus two times the acceleration times displacement.

I got 0= V initial squared + 2(-10.0)(34.04)
V initial squared = 680.6881
V initial = 26.09
It didn't work... or maybe I'm doing it wrong...

FlyingDutchman
09-13-2009, 07:03 PM
I tried using Velocity Final squared is equal to Velocity Inital Squared plus two times the acceleration times displacement.

I got 0= V initial squared + 2(-10.0)(34.04)
V initial squared = 680.6881
V initial = 26.09
It didn't work... or maybe I'm doing it wrong...

hmm well your lookin for the max velocity he can go right?

88 km/h = 24.444444444 m/s

so first you figure out how much distance his reaction takes (0.75s)

d = (v)(t) -> d = (24.4)(0.75) = 18.3m til he hits the brakes leaving 19.7m to stop yes?

Knipps
09-13-2009, 07:11 PM
Why are you guys doing this kid's HW?

FlyingDutchman
09-13-2009, 07:15 PM
Why are you guys doing this kid's HW?

i ... dont ... know .... lol but now i wanna know the right answer

ljstella
09-13-2009, 07:20 PM
I came up with 88km/h as something else in m/s...

I multiplied 88 by 60 to get km/s and then divided by 1000 (that's my error... I should have multiplied by 1000 to get m/s) but then I get something ridiculously huge... if I divide, I get 5.28 m/s and then multiply that by .75 and subtract the product from 38.0 to get 34.04m as the distance from when he starts to brake...

FlyingDutchman
09-13-2009, 07:33 PM
I came up with 88km/h as something else in m/s...

I multiplied 88 by 60 to get km/s and then divided by 1000 (that's my error... I should have multiplied by 1000 to get m/s) but then I get something ridiculously huge... if I divide, I get 5.28 m/s and then multiply that by .75 and subtract the product from 38.0 to get 34.04m as the distance from when he starts to brake...

km/h -> m/s

x1000 then / 3600

LTb1ow
09-13-2009, 07:34 PM
So this proved you all suck at physics.

ljstella
09-13-2009, 07:36 PM
km/h -> m/s

x1000 then / 3600

why 3600?

WildBillyT
09-13-2009, 07:41 PM
6.023 x 10^23

FlyingDutchman
09-13-2009, 07:41 PM
So this proved you all suck at physics.

why?

ljstella
09-13-2009, 07:45 PM
6.023 x 10^23

Physics, not chemistry! Avogadro's number isn't going to help me!!

madness410
09-13-2009, 08:54 PM
he would end up in jail. speed limit is only 65.

jims69camaro
09-14-2009, 06:32 AM
So this proved you all suck at physics.

:nod:

why?

because you haven't come up with the right formula yet, which means if you do hit the answer (by some means, perhaps the russian's beaming method) it will be by pure luck, not because you have applied the right formula.

that's all physics is unless you are among the elite - then you get to try your own formulas and spend years proving/disproving them, supported, of course, by educational grants from the government and wealthy privateers... applying the right formula.

WildBillyT
09-14-2009, 08:42 AM
So this proved you all suck at physics.

No, the answer is 19.67 m/s. I posted Avegadro's number because I felt like busting balls- as he should really do his own homework so he doesn't get boned on a test.

I got it the same way 97Camaro did. 18.33m covered until the brakes are applied, 19.67m of distance left, equation will zero out when the initial velocity is ~19.83m/s. The only "ball buster" is how far they want the car to be from the wall to be considered "not hit". 0 is not really right IMO, because they would be exactly touching.

Tsar
09-14-2009, 10:09 AM
No, the answer is 19.67 m/s. I posted Avegadro's number because I felt like busting balls- as he should really do his own homework so he doesn't get boned on a test.

I got it the same way 97Camaro did. 18.33m covered until the brakes are applied, 19.67m of distance left, equation will zero out when the initial velocity is ~19.83m/s. The only "ball buster" is how far they want the car to be from the wall to be considered "not hit". 0 is not really right IMO, because they would be exactly touching.So 59 is NOT the answer? :?:

WildBillyT
09-14-2009, 10:13 AM
So 59 is NOT the answer? :?:

No?

FlyingDutchman
09-14-2009, 10:13 AM
because you haven't come up with the right formula yet, which means if you do hit the answer (by some means, perhaps the russian's beaming method) it will be by pure luck, not because you have applied the right formula.

that's all physics is unless you are among the elite - then you get to try your own formulas and spend years proving/disproving them, supported, of course, by educational grants from the government and wealthy privateers... applying the right formula.

orly? i used my cheat sheet from physics last year with all the kinematic (http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/1DKin/U1L6a.html) equations and those are the 'right' formulas

Tsar
09-14-2009, 10:17 AM
No?
Oh man, I'm so disappointed!

WildBillyT
09-14-2009, 10:27 AM
orly? i used my cheat sheet from physics last year with all the kinematic (http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/1DKin/U1L6a.html) equations and those are the 'right' formulas

LOL. I used the exact same website. Couldn't remember most of them.

LTb1ow
09-14-2009, 10:28 AM
Did not figure in speed of light, fail.


:wink:

WildBillyT
09-14-2009, 10:31 AM
Did not figure in speed of light, fail.


:wink:

If you want we can help you determine the electrical resistance of a dead fuel pump.

LTb1ow
09-14-2009, 10:38 AM
Prob pretty high. I would guess more than enough to melt those puny wires that feed it.

sweetbmxrider
09-14-2009, 10:53 AM
wait, what motor?

LTb1ow
09-14-2009, 11:31 AM
http://www.capeconcrete.com/products/images/new_jersey_barrier_3000.jpg

Jersey barrier. Gotta love it.

ljstella
09-14-2009, 12:16 PM
Just figured I'd let you know, I figured out how to solve it :)

sweetbmxrider
09-14-2009, 12:26 PM
Jersey barrier. Gotta love it.

ya, good for wallrides and quarter pipe stuff 8-)

FlyingDutchman
09-14-2009, 12:31 PM
and crashing smart cars (http://www.youtube.com/watch?v=ju6t-yyoU8s)

ljstella
09-14-2009, 06:07 PM
So, I had to use the quadratic equation, which I didn't think about, and the answer ended up being 75.852 km/h

LTb1ow
09-14-2009, 06:11 PM
Wootz for complex numbers!

MyFirstZ
09-14-2009, 07:28 PM
what kind of brakes was he using. . . baer, brembo, stock? # of pistons in the caliper? friction material, rotor size (friction surface). . . i need things that matter here, number wont help

jims69camaro
09-18-2009, 10:52 AM
orly? i used my cheat sheet from physics last year with all the kinematic (http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/1DKin/U1L6a.html) equations and those are the 'right' formulas

thank you so much for that website. i have my high school textbook around here somewhere, but i have been unable to find it. that website kicks ass.