Length of a Collision

[MedPR]

The front 1.2m of a 1400kg car is designed as a "crumple-zone" that collapses to absorb the shock of a collision. If a car traveling at 25m/s stops uniformly in 1.2m, how long does the collision last?

A. 4.8*10^-2
B. 9.6*10^-2

Answer is B because vt=x, where v=25m/s and x=1.2m, but you have to multiple x by 2. Why do you have to multiply x by 2?

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[chiddler]

Think how the collision is occuring.

Not very easy to draw, but I hope this makes some sense.

The front of the car must collapse 1.2 meters before travelling 1.2 meters forward. So we've just doubled the amount of distance in which the impact occurs.

 

[milski]

vt = x would be how much time it would take if it was moving at constant speed v.

For uniform acceleration:

1.2=25*t+a/2*t^2
0=25+t*a

ta=-25
a=-25/t
1.2=25*t-t^2*25/(2t)=25t-25t/2=25t/2
t=2.4/25
t=0.096

There may be some shortcut formula with velocity and time only but I don't have that memorized.

 

[MedPR]

Think how the collision is occuring.

Not very easy to draw, but I hope this makes some sense.

The front of the car must collapse 1.2 meters before travelling 1.2 meters forward. So we've just doubled the amount of distance in which the impact occurs.

Oh so the problem is saying it crumples and then travels another 1.2m before it stops? That doesn't make sense to me conceptually :/

 

[chiddler]

Oh so the problem is saying it crumples and then travels another 1.2m before it stops? That doesn't make sense to me conceptually :/

well it made sense 10 minutes ago lol.

i'm not sure anymore.

glad milski is around.

 

[chiddler]

1.2=25*t+a/2*t^2
0=25+t*a

what is this?

 

[milski]

what is this?

v=v0+a*t

a and t are unknown, so you need two equations.

Like I said, there is a formula relating only v, v0, x and t but I don't remember it. It's not vx=t though.

 

[chiddler]

v=v0+a*t

a and t are unknown, so you need two equations.

Like I said, there is a formula relating only v, v0, x and t but I don't remember it. It's not vx=t though.

What did you do differently from medpr? What was his mistake? It looks like you're doing the same although using different formulas.

 

[milski]

What did you do differently from medpr? What was his mistake? It looks like you're doing the same although using different formulas.

We're doing the same, in a sense that we are writing equations which describe the motion of the car.

He used vt=x. That's the displacement for constant velocity and is not really correct in this case since the car is stopping.

I used the formulas for distance with uniform acceleration and velocity with uniform acceleration.

 

[chiddler]

We're doing the same, in a sense that we are writing equations which describe the motion of the car.

He used vt=x. That's the displacement for constant velocity and is not really correct in this case since the car is stopping.

I used the formulas for distance with uniform acceleration and velocity with uniform acceleration.

I wish I figured this out without you telling me =/

thanks.

 

[milski]

I wish I figured this out without you telling me =/

thanks.

No problem, I am capable of witholding information at arbitrary times.

 

[SaCkO]

Guys from the way i read the question this seems like a collision question with impulse involved.
Impulse=F*t
F=ma
impulse is mv2-mv1=ma*t
in order to get a we use Vf^2=Vi^2+2ax
we get a=-260
then V2-V1=-260*t
t=25/260=9.6*10^-2

 

[milski]

That works too. There's more than one way to skin a dog.

 

[MedPR]

Ah I see, thanks everyone!

 

[MT Headed]

The one and only one equation you are looking for is: X = Vavg T

Vavg is the average velocity of the interaction, or (25 + 0)/2 = 12.5

So the answer is T = X / Vavg = 1.2 / 12.5 = a little bit less than 0.1

 

[MedPR]

The one and only one equation you are looking for is: X = Vavg T

Vavg is the average velocity of the interaction, or (25 + 0)/2 = 12.5

So the answer is T = X / Vavg = 1.2 / 12.5 = a little bit less than 0.1

You know, back when I was doing the TBR kinematics chapter 3 months ago I always forgot about this equation and I always wanted to bang my head against the desk when I did all the problems the long way then remembered it.

Thanks for reminding us

 

[pfaction]

I did it in multiple steps...

Vsq = Vosq+2(a)(d)
0 = 625 + 2.4(a)
a = 250
v=v0+at
0=25+250(t)
25/250 = 1/10 = 0.1
9.6*10^-2 is 0.096 = 0.01

 

[milski]

The one and only one equation you are looking for is: X = Vavg T

Vavg is the average velocity of the interaction, or (25 + 0)/2 = 12.5

So the answer is T = X / Vavg = 1.2 / 12.5 = a little bit less than 0.1

Sweet! Short, simple and even works for non-uniform acceleration!

 

[MedPR]

Sweet! Short, simple and even works for non-uniform acceleration!

I think this one only works for uniform acceleration.

 

[pfaction]

Does my answer work for anyone? Does it make sense?

 

[MedPR]

Does my answer work for anyone? Does it make sense?

Yes it works and makes sense. Thank you

 

[pfaction]

Thanks. As you know I got severely raped on AAMC10 and I want to make sure my stuff is good.

 

[MedPR]

Thanks. As you know I got severely raped on AAMC10 and I want to make sure my stuff is good.

I did not know that. What did you get?

 

[pfaction]

9 (so many reading mistakes dude. I should have gotten 10 or 11.)
7 (verbal is an amplified achilles heel)
13 (again, I made reading mistakes, and second guessed myself)
TPR averages:
9
7
14

To summarize how bad my reading skills are: I misread "freeze" for "melt" and answered "boiling point". I looked at my scrap paper and literally said out loud, "what the ****?"

 

[milski]

I think this one only works for uniform acceleration.

The Vavg works for anything. How do you find the average velocity is a whole separate story. It's not (Vf+Vi)/2 in that case.

 

[MedPR]

The Vavg works for anything. How do you find the average velocity is whole separate story though. It's not (Vf+Vi)/2 in that case.

Oh that's right, the (Vf+Vi)/2 only works for constant acceleration. Thanks again

 

[Ineedhopenow]

The one and only one equation you are looking for is: X = Vavg T

Vavg is the average velocity of the interaction, or (25 + 0)/2 = 12.5

So the answer is T = X / Vavg = 1.2 / 12.5 = a little bit less than 0.1

My mind = blown at the simplicity...

Thanks MT!!!

 

[Class1P]

I found the K.E then divided it by the distance to get the Force. Then I found the change in momentum and divided it by the force which equals the time. it wasn't easy to do without a calculator so I am sure one of your other ways is better.

KE = F*d

Divide by d = F

t = mv/f