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Old 09-13-2009, 05:33 PM   #1
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Question High School Physics

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.
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Old 09-13-2009, 05:50 PM   #2
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Old 09-13-2009, 05:58 PM   #3
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threve
...what?
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Old 09-13-2009, 06:01 PM   #4
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57.
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Old 09-13-2009, 06:04 PM   #5
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57.
No...
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Old 09-13-2009, 06:09 PM   #6
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Yes
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Old 09-13-2009, 06:13 PM   #7
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Yes
The website rejected the answer. So, I guess not.

I hate WebAssign
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Old 09-13-2009, 06:14 PM   #8
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wait does he hit the brakes with 38.0m to go or is 38.0m then he hits the brakes?
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Old 09-13-2009, 06:15 PM   #9
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Quote:
Originally Posted by 97camaro View Post
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.

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Old 09-13-2009, 06:16 PM   #10
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Ok try 58.
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Old 09-13-2009, 06:17 PM   #11
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Quote:
Originally Posted by Tru2Chevy View Post
When he sees the barrier he is 38m away, then .75 seconds later he actually applies the brakes.

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ok thought soo
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Old 09-13-2009, 06:19 PM   #12
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Ok try 58.
Not that either...
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Old 09-13-2009, 06:19 PM   #13
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59?
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Old 09-13-2009, 06:21 PM   #14
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59?
...are you just guessing?
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Old 09-13-2009, 06:22 PM   #15
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idk tsar...I came up with 59.4km/h
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Old 09-13-2009, 06:23 PM   #16
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Originally Posted by ljstella View Post
...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.
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Old 09-13-2009, 06:24 PM   #17
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w00t
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Last edited by FlyingDutchman; 09-13-2009 at 06:47 PM.
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Old 09-13-2009, 06:24 PM   #18
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at least it didn't come in as 75
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Old 09-13-2009, 06:25 PM   #19
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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...
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Old 09-13-2009, 06:27 PM   #20
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How many more do you have left?
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Old 09-13-2009, 06:28 PM   #21
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Quote:
Originally Posted by Tsar View Post
How many more do you have left?
59...
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Old 09-13-2009, 06:34 PM   #22
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Originally Posted by 97camaro View Post
59...
No, you lie.
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Old 09-13-2009, 06:36 PM   #23
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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
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Last edited by FlyingDutchman; 09-13-2009 at 06:46 PM.
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Old 09-13-2009, 06:39 PM   #24
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Originally Posted by 97camaro View Post
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.
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Old 09-13-2009, 06:47 PM   #25
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Quote:
Originally Posted by Tsar View Post
That's not correct.
how so?
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