Advanced One Dimensional Kinematic Problem

[LexRunner]

A car is traveling at 25 m/s when it runs off the road and hits a utility pole. The car stops instantly, but the driver continues to move forward at 25 m/s. The airbag starts from rest with a constant acceleration from a distance of 50 cm away from the driver and makes contact with him in 9 ms. What is the acceleration of the airbag?

My answer was a = 1234.568 m/s, but its wrong.

These were what I believe to have been the known quantities:
v_i = 0
a = ?
t = 0.009 s
d = 0.050 m

I used this equation to solve the problem:
d = v_it + 1/2*a*t²

What am I doing wrong and what is the right answer?
Thanks

 

[SteamKing]

How many meters is 50 cm equivalent to?

 

[nrqed]

A car is traveling at 25 m/s when it runs off the road and hits a utility pole. The car stops instantly, but the driver continues to move forward at 25 m/s. The airbag starts from rest with a constant acceleration from a distance of 50 cm away from the driver and makes contact with him in 9 ms. What is the acceleration of the airbag?

My answer was a = 1234.568 m/s, but its wrong.

These were what I believe to have been the known quantities:
v_i = 0
a = ?
t = 0.009 s
d = 0.050 m

I used this equation to solve the problem:
d = v_it + 1/2*a*t²

What am I doing wrong and what is the right answer?
Thanks

The driver was moving, so the airbag does not have to travel the full 50 cm. The bag and the driver must travel that distance so the airbag will travel a smaller distance.

By the way, the answer will be in $m/s^2$, not $m/s$.

 

[LexRunner]

The driver was moving, so the airbag does not have to travel the full 50 cm. The bag and the driver must travel that distance so the airbag will travel a smaller distance.

By the way, the answer will be in $m/s^2$, not $m/s$.

So you're saying my other quantities are correct, except the quantity for distance because the driver will still be moving forward while the airbag is coming out so he distance would be less than 50 cm. So if I found the right distance, and plugged it into the equation, I would gt the right answer?

 

[LexRunner]

So to find the distance the airbag traveled, I would have to take 50 cm subtracted by the distance the man traveled. These would by my known quantities:
V(initial): 25 m/s
V(final): 0 m/s (because he hits the airbag)
T: 0.009 s (convert 9 ms to sec)
D: ?

So I would have to plug his into find the distance the man travelled, take 50 cm subtracted by the value of D to find the distancetthe airbag traveled and plug the result into the original equation to find the acceleration of the airbag.

 

[BvU]

Apparently you are working in a reference frame attached to the car. Which is standing still against the utility pole when the airbag takes off.

So to find the distance the airbag traveled, I would have to take 50 cm subtracted by the distance the man traveled.

is indeed correct. Please calculate this distance, which under point 1. has been called d, and show.
Now the airbag.

The airbag starts from rest with a constant acceleration

so you apply the relevant equation (so conveniently listed under 2. relevant equations. In the template.)

d = v_it + 1/2*a*t²

What is v_i if it starts from rest ?

This gives you the so desired a. Note that v_f is not zero. The man hits the bag with 25 m/s for the man and v_f (which should be considerable) in the other direction for the bag. It's not a sandbag, it's an airbag.

Ceterum censeo that this kind of exercises is didactically irresponsible: there is nothing realistic about it, it's full of unsound physics, etc. etc.

Nevertheless: welcome to PF, Lex ! Not your fault what I fulminate against in the last paragraph!

 

[dean barry]

You know how far the man travelles (in metres) in 0.009 seconds, its : v * t = 25.0 * 0.009 = 0.225 m
Subtract that from 0.5 m to get the distance traveled by the airbag (s), then :
a = ( s - ( u * t ) ) / ( ½ * t ² )
( t = 0.009 s, u = 0.0 )

 

[BvU]

Thanks, Dean. But the intention was that Lex should do and show the work...