Acceleration car crash g's experienced?

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Homework Help Overview

The discussion revolves around calculating the average acceleration experienced by a driver during a car crash, expressed in multiples of g (where g = 9.8 m/s²). The original poster presents a scenario where a car's velocity decreases from 26 m/s to 0 m/s over a specified time, initially stated as 15 seconds, which raises questions about the accuracy of the time frame given the context of a crash.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the correct application of the kinematic equation for acceleration and question the validity of the time duration provided. There is uncertainty about the initial and final velocities, and participants explore how to set up the problem correctly to find the acceleration in terms of g's.

Discussion Status

The conversation is ongoing, with participants providing hints and guidance on how to rearrange the equation for acceleration. There is a focus on clarifying the time duration and ensuring the correct values are used in calculations. Some participants are attempting to recalculate based on the corrected time frame, while others are checking assumptions about the problem setup.

Contextual Notes

There is confusion regarding the time variable, with a suggestion that it may be 0.15 seconds instead of 15 seconds. This discrepancy significantly impacts the calculations being discussed. Additionally, participants note that the answer choices do not include decimal points, which adds to the complexity of the problem.

physicsgurl12
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acceleration car crash g's experienced??

Homework Statement



Acceleration is sometimes expressed in multiples of g where g=9.8m/s^2 is the acceleration due to the Earth's gravity. In a car crash the cars velocity may go from 26m/s to 0m/s in 15s. how many g"s are experienced on average by the driver??
a, 23- i know this is wrong
b, 18g
c,22g
d,23g

Homework Equations



we have a, vi, vf and t
so something like v=v0+at

The Attempt at a Solution


v=26m/s+ 9.8m/s^2*15s=173??
 
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The general equation you have is correct, yes. However, look what you know you have. Do you know the final velocity? How about the initial velocity? Does the problem give you the time it takes for the change in velocity?

Also, look at the given problem in your book or on the assignment. The number seems a little off for time. Are you sure it is not 0.15s? 15s would be more like someone stepping on the brakes of a car, rather than an impact.
 


Ignea_unda said:
The general equation you have is correct, yes. However, look what you know you have. Do you know the final velocity? How about the initial velocity? Does the problem give you the time it takes for the change in velocity?

Also, look at the given problem in your book or on the assignment. The number seems a little off for time. Are you sure it is not 0.15s? 15s would be more like someone stepping on the brakes of a car, rather than an impact.

o haha. i guess i typed the time wrong. so that helps a bit but i redid it and got 27.
 


Okay, look at the first half of the hint I gave. Did you rearrange the equation based on what you know? What does it look like when you do that?
 


okay well. I am not really sure what the question is asking me to find.It says g's. i know final and initial velocity and the time it takes to change. but how do i set that up?
 


physicsgurl12 said:

Homework Equations



we have a, vi, vf and t
so something like v=v0+at

Yep. That's the equation you need.
However, you don't have "a" yet.
"a" is what you need to calculate with this formula.

v0 is the same as vi.
And v is the same as vf.

So you have:
vf = vi + a t


physicsgurl12 said:

The Attempt at a Solution


v=26m/s+ 9.8m/s^2*15s=173??

In this formula you substituted g=9.8 for the acceleration.
But that's not right.
"g" is not the acceleration.

And v is not 173 after 15 seconds.
v is vf.

Can you substitute the final velocity vf, and then calculate "a" using your formula?
 


physicsgurl12 said:
okay well. I am not really sure what the question is asking me to find.It says g's. i know final and initial velocity and the time it takes to change. but how do i set that up?

Yes, it says g's.
This means that after you have calculated "a", you need to divide it by "g" to get an answer in g's.
 


okay so then my equation is vf-vi/t=a?? because then you get -17.68 which is pretty close to 18 but its negative.
 


physicsgurl12 said:
okay so then my equation is vf-vi/t=a?? because then you get -17.68 which is pretty close to 18 but its negative.

It is negative because the speed is decreasing to zero.
However, the acceleration acting on the driver is the magnitude of this acceleration and as such positive.

And you should check your numbers.
If I use vi = 26 m/s
and t = 15 s
then I'm getting a = -26 / 15 / 9.8 = -0.1768 g

So you seem to have a couple of decimal points in the wrong place...
 
  • #10


okay but there's no decimal points in the answer choices??
 
  • #11


physicsgurl12 said:
okay but there's no decimal points in the answer choices??

First, are you sure your numbers 26 m/s and 15 s are correct?
You remarked earlier that you made a typo, but you did not clarify.
Is it perhaps .15 s? Which is 0.15 s.

Then, when you write: "b, 18g"
Could that perhaps be: "b ,18g" which is actually "b 0.18g"?
 
  • #12


I like Serena said:
First, are you sure your numbers 26 m/s and 15 s are correct?
You remarked earlier that you made a typo, but you did not clarify.
Is it perhaps .15 s? Which is 0.15 s.

Then, when you write: "b, 18g"
Could that perhaps be: "b ,18g" which is actually "b 0.18g"?

yes it was .15s but there's still no decimals in the answers
 
  • #13


physicsgurl12 said:
yes it was .15s but there's still no decimals in the answers

So redo the calculation with t = 0.15 s?
 

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