Acceleration car crash g's experienced?

In summary: Just to be sure: vf = vi + a tSo a = (vf - vi) / tSoa = (0 - 26) / 0.15 / 9.8 = -177.41935483870968Now, I have no idea what the correct answer is, but that it seems in the ballpark of -18 gIn summary, the problem involves calculating the acceleration experienced by a driver in a car crash, given the initial and final velocities, and the time it takes for the change in velocity. The correct formula to use is vf = vi + at, where "a" represents the acceleration. After rearranging the formula, the acceleration is calculated to be -177
  • #1
physicsgurl12
184
0
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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  • #2


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.
 
  • #3


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.
 
  • #4


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?
 
  • #5


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?
 
  • #6


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?
 
  • #7


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.
 
  • #8


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.
 
  • #9


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?
 

Related to Acceleration car crash g's experienced?

1. What is the definition of "g" in a car crash?

The term "g" refers to acceleration due to gravity, which is equal to 9.8 meters per second squared (m/s²). In a car crash, the acceleration experienced by the vehicle and its passengers is often measured in terms of "g-forces."

2. How is the acceleration experienced in a car crash measured?

Acceleration in a car crash is typically measured in terms of "g-forces," which is a unit of measurement that compares the force experienced to that of Earth's gravity. For example, if a car experiences 3 g's in a crash, it means that the force experienced is three times that of Earth's gravity.

3. What is a "high" g-force in a car crash?

A "high" g-force in a car crash is any force that is greater than what a person typically experiences during normal everyday activities. This can vary depending on the individual, but generally, anything above 5 g's is considered high and can result in serious injuries.

4. What factors can affect the g-forces experienced in a car crash?

The g-forces experienced in a car crash can be affected by various factors, including the speed of the vehicles involved, the angle of impact, the size and weight of the vehicles, and the use of safety features such as seatbelts and airbags. The type of collision, whether it be a head-on or side impact, can also impact the g-forces experienced.

5. How can high g-forces in a car crash be reduced?

The best way to reduce the g-forces experienced in a car crash is to practice safe driving habits and follow traffic laws. Additionally, using safety features such as seatbelts and airbags, as well as choosing a vehicle with good crash test ratings, can also help reduce the impact of g-forces during a crash.

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