• Support PF! Buy your school textbooks, materials and every day products Here!

Car Crash! Motion in 1 dimension Question

  • Thread starter Vroc
  • Start date
  • #1
30
0

Homework Statement



During a car accident, a vehicle with an intial velocity of 100km/h hits a concrete wall. The "crumple zone" in the front of the vehicle is a space that makes up the engine compartment
that is designed to allow the passenger compartment to continue forward a distance of 1.20m under a constant acceleration (negative) before it comes to a stop. What is the average velocity for the "crash"? Calculate the time taken to decelerate to a stop.

Vi= 27.8m/s (converted from 100km/h)
Vf= ?
Distance= ?
Acceleration= ?
Time=? Required: Average Velocity, Time

Homework Equations


Vav= V1 + V2 divided by 2
Vf= V1 + at
Vf(sq)= v1(sq) + 2ad


The Attempt at a Solution


I heard that you can't just take 0 as the final velocity and in that case, I have clue as to where to start this question. I need three variables to get anywhere but it looks like I only have two.
 

Answers and Replies

  • #2
537
1
if it stops, then the final velocity is zero
 
  • #3
30
0
Okay. Now how do I put the "crumple zone" portion into effect? I know that if no acceleration is given, It's 9.8 but in this case there's that crumple zone thing. How would I calculate the time taken for the car to decelerate to a stop?
 
  • #4
30
0
if it stops, then the final velocity is zero
Okay, so that gives me two variables. I need three to figure out the time taken to decelerate to a stop.
 
  • #5
537
1
All the question is saying is that something started off with some initial velocity, then decelerates over some distance until it stops.

You're going to need to solve for the acceleration to finish this problem. I suggest using the third equation you referenced.

Also, 9.8m/s^2 would be the acceleration due to gravity. In this situation the object (the passenger compartment) is only moving in the x direction, which is not going up or down. So there's no gravity here.
 
  • #6
30
0
All the question is saying is that something started off with some initial velocity, then decelerates over some distance until it stops.

You're going to need to solve for the acceleration to finish this problem. I suggest using the third equation you referenced.

Also, 9.8m/s^2 would be the acceleration due to gravity. In this situation the object (the passenger compartment) is only moving in the x direction, which is not going up or down. So there's no gravity here.
I only have two variables though. I need to know the distance to figure out using the third equation I referenced. Should I use 1.20m as the distance?
 
  • #7
537
1
I only have two variables though. I need to know the distance to figure out using the third equation I referenced. Should I use 1.20m as the distance?
yes, the problem tells you that "the passenger compartment to continue forward a distance of 1.20m under a constant acceleration (negative) before it comes to a stop"
 
  • #8
30
0
yes, the problem tells you that "the passenger compartment to continue forward a distance of 1.20m under a constant acceleration (negative) before it comes to a stop"
Thanks for the help so far. But just to be clear, After I get my answer I should change it to a negative correct?

Thanks again,
Vroc
 
  • #9
537
1
well, actually if you use that third equation, the acceleration should end up negative

because if the final velocity is negative, then you have 0 = v0 + 2ad

so a = -v0/2d

if you keep track of all the proper signs, then the answer will end up with the right sign at the end

for instance, if the object starts at 0 m/s and then goes to some positive final velocity, then you will find that a = vf/2d, a positive number

we could also define our coordinate system so that the object is moving in the negative x direction, starting from the origin. If we did this, then we'd have to change the distance traveled to a negative number and also the initial velocity would have to become negative.

If you work it out like I did above, you should find that the acceleration will turn out to be positive. But the important thing to note is that the acceleration in this situation will be in the opposite direction of the initial velocity, because the object is *slowing down*. It won't always be *negative*, but if the object is slowing down, the acceleration will be in the *opposite direction*, meaning that it will have the opposite sign of the initial velocity.
 
  • #10
30
0
well, actually if you use that third equation, the acceleration should end up negative

because if the final velocity is negative, then you have 0 = v0 + 2ad

so a = -v0/2d

if you keep track of all the proper signs, then the answer will end up with the right sign at the end

for instance, if the object starts at 0 m/s and then goes to some positive final velocity, then you will find that a = vf/2d, a positive number

we could also define our coordinate system so that the object is moving in the negative x direction, starting from the origin. If we did this, then we'd have to change the distance traveled to a negative number and also the initial velocity would have to become negative.

If you work it out like I did above, you should find that the acceleration will turn out to be positive. But the important thing to note is that the acceleration in this situation will be in the opposite direction of the initial velocity, because the object is *slowing down*. It won't always be *negative*, but if the object is slowing down, the acceleration will be in the *opposite direction*, meaning that it will have the opposite sign of the initial velocity.
V2(sq) = vi(sq) + 2ad
0 = 27.8(sq) + 2 x a x 1.2
move 27.8(sq) over and than square it. 2 x 1.2 =2.4. It was then moved over.
772.84 / 2.4 = a
a = 322.02m/s, This seems ridiculous does it not? Ahhh, I don't know where I made the mistake. Even if the 27.8 is not squared the answer seems weird as well. (27.8/2.4 = 11.6m/s)
 
  • #11
537
1
V2(sq) = vi(sq) + 2ad
0 = 27.8(sq) + 2 x a x 1.2
move 27.8(sq) over and than square it. 2 x 1.2 =2.4. It was then moved over.
772.84 / 2.4 = a
a = 322.02m/s, This seems ridiculous does it not? Ahhh, I don't know where I made the mistake. Even if the 27.8 is not squared the answer seems weird as well. (27.8/2.4 = 11.6m/s)
the answer may seem ridiculous, but have you ever watched a car hit a concrete wall? The deceleration is pretty fast.

for example, try finding how long it takes now, should be a really short amount of time
 
  • #12
30
0
Okay, I'm glad I was right to square it. I was actually contemplating putting down the 11.6 instead because it was more reasonable. So 322.02m/s is the answer huh? I think I'm beating a dead horse at this point but you said the answer should end up as negative.I just don't want to end up loosing an easy checkmark on this question. Also, that number is pretty Weird considering there was a dragster question earlier and it went 400m and travelled at at an acelleration rate of 7.84m.
 
  • #13
537
1
well when you moved the 27.8(sq) over, you should end up with it negative, because you are subtracting it from both sides of the equation

vf2 = v2i + 2ad
vf2 - v2i = 2ad
(vf2 - v2i)/2d = a

since vf = 0, you should get a negative answer for a

and I'd suggest finding some crash test videos on youtube to get an appreciation for the deceleration caused by crashing into something at a relatively high speed
 
  • #14
3,732
414
a = 322.02m/s, This seems ridiculous does it not?
The only ridiculous think is using m/s as unit for acceleration.
m/s is for velocity or speed. Acceleration is measured in m/s^2 or m/s/s.
You should be more careful with units. They are important.
Otherwise your statements become meaningless, like, for example this one:
"it went 400m and travelled at at an acelleration rate of 7.84m".

Going back to the above value (assuming that is in m/s^2), it is not weird, for the data given in the problem.
It may be unrealistic, as in reality the acceleration is not constant during the crash.
 
  • #15
537
1
Proper units is definitely something you should be careful with, that's an easy thing to lose points on. I can guarantee that you will lose points if you neglect to put the right units.
 
  • #16
30
0
well when you moved the 27.8(sq) over, you should end up with it negative, because you are subtracting it from both sides of the equation

vf2 = v2i + 2ad
vf2 - v2i = 2ad
(vf2 - v2i)/2d = a

since vf = 0, you should get a negative answer for a

and I'd suggest finding some crash test videos on youtube to get an appreciation for the deceleration caused by crashing into something at a relatively high speed
But when you square a number with an even exponent(vi2) It becomes positive. And as you predicted, the time seems even more crazy. 0.086 seconds Ahah, Will do. :tongue2:
 
Last edited:
  • #17
537
1
But when you square a number with an even exponent(vi2) It becomes positive. And as you predicted, the time seems even more crazy. 0.086 seconds Ahah, Will do. :tongue2:
there's no squaring of negative numbers going on

0 - vf2 is going to be negative

that means that if it gets divided by a positive number (and 2d is a positive number) then the acceleration will be negative
 

Related Threads for: Car Crash! Motion in 1 dimension Question

  • Last Post
Replies
0
Views
1K
  • Last Post
2
Replies
33
Views
11K
  • Last Post
Replies
8
Views
856
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
6
Views
6K
Top