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Physics Problem (Acceleration / Braking)

P

PCMarine

Diamond Member
Hello ATOT, it would be great if you guys could help me out with this physics problem:

The driver of a car going 90.0 km/h suddenly sees the lights of a barrier 40.0 m ahead. It takes the driver 0.75 s to apply the brakes, and the average acceleration during braking is -10.0 m/s^2.

(The actual problem): What is the maximum speed at which the car could be moving and not hit the barrier 40.0 m ahead? Assume that the acceleration rate doesn't change."
Click to expand...

My physics teacher said that the answer is somewhere between 20 and 24.
 
first you need to convert everything to the same units (i used m/s),

90 km/hr -> 25 m/s
40 m -> 40 m
.75 s -> .75 s
-10 m/s^2 -> -10 m/s^2

now, we just need to figure out at what distance from the barrier did the driver apply the brakes,

(.75s)(25 m/s)=18.75 m.

therefore, there is 40-18.75=21.25m of road left. using this info and the fact that the driver brakes at -10m/s^2 we can use the following kinematic equation,

vfinal^2=vinitial^2+2*a*d,

where vfinal=0 (max speed to stop right at the barrier), vinitial is what we're solving for, a=-10, and d=21.25.

plugging this all in and solving, i get,

vinitial = 20.62 m/s approx.
 
Originally posted by: maziwanka
first you need to convert everything to the same units (i used m/s),

90 km/hr -> 25 m/s
40 m -> 40 m
.75 s -> .75 s
-10 m/s^2 -> -10 m/s^2

now, we just need to figure out at what distance from the barrier did the driver apply the brakes,

(.75s)(25 m/s)=18.75 m.

therefore, there is 40-18.75=21.25m of road left. using this info and the fact that the driver brakes at -10m/s^2 we can use the following kinematic equation,

vfinal^2=vinitial^2+2*a*d,

where vfinal=0 (max speed to stop right at the barrier), vinitial is what we're solving for, a=-10, and d=21.25.

plugging this all in and solving, i get,

vinitial = 20.62 m/s approx.
Click to expand...


It's not really that simple though, because the initial calculation ((.75s)(25 m/s)=18.75 m.) Depends upon the starting speed which we are looking for. You have to propogate that down as well.
 
The actual answer (accounting for a change in the initial speed):

Basically the same as the solution outlined above, but you have to carry the initial reaction time distance down:

(.75s)(Vi) = reaction distance

using kinematics

Vf^2=Vi^2+2*a*d,

Vf=0 , a=-10, and d=40-reaction distance= 40-(.75s)(Vi)

0^2=Vi^2+2(-10)(40-(.75s)(Vi))

Vi^2+15Vi-800=0 , solve for quadratic

Vi = 21.76 m/s = 78.34 km/hr
 
i knew the question was weird. i assumed (idiotically) that he was going 25 m/s (which the original post stated) traveled some distance and then from that distance on we were supposed to see at what speed the driver could have gone (at the most) to avoid hitting the barrier.

absolutdealage, your answer is definately right if the question is to determine the speed he could go to avoid hitting that barrier assuming his reaction time is .75 sec (disregarding that 25 m/s in the beginning) (makes more sense).

sweet.
 
Originally posted by: funboy6942
WOW are you studing to be a peace officer or crash inspector for an insurance company???
Click to expand...

No, this is first semester physics... could be any number of degrees: engineering, physics, math, etc.
 
Wow, thanks a lot guys. I was kinda expecting "WTF do you own g*d damned work!", but you guys really helped me out!

Thanks!
 
Originally posted by: AbsolutDealage
The actual answer (accounting for a change in the initial speed):

Basically the same as the solution outlined above, but you have to carry the initial reaction time distance down:

(.75s)(Vi) = reaction distance

using kinematics

Vf^2=Vi^2+2*a*d,

Vf=0 , a=-10, and d=40-reaction distance= 40-(.75s)(Vi)

0^2=Vi^2+2(-10)(40-(.75s)(Vi))

Vi^2+15Vi-800=0 , solve for quadratic

Vi = 21.76 m/s = 78.34 km/hr
Click to expand...

Parden me for being retarded, but how would "2(-10)(40-(.75s)(Vi))" factor out into 15V - 800?
 
Originally posted by: BladeWalker
ATOT became a tutoring session? You guys are too generous.
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it's not a tutoring session, it's an answer-giving session. hints go a lot further towards learning than just giving someone the answer.
 
Originally posted by: Triumph
Originally posted by: BladeWalker
ATOT became a tutoring session? You guys are too generous.
Click to expand...

it's not a tutoring session, it's an answer-giving session. hints go a lot further towards learning than just giving someone the answer.
Click to expand...

IMO, if someone is asking on this board... they either just want the answer or they want to check their work. If they really just want the answer, they are going to get it one way or another, so why not just show the process and give the answer? If you're not going to do the work yourself, then what's the difference?

He could just as easily have gone to talk to the prof/TA about it, and they would have given hints out. That's not what he wanted. He just wanted "the fish".
 
Because you should discourage that kind of activity. Just because they'll get it somewhere else isn't a valid argument. If he found that he couldn't find anywhere to copy the answers from, maybe he'd then do the work himself, and be a smarter person for it.
 
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