Sports car stopping distance: average acceleration from 95 km/h over 44 m

[slayerdeus]

A sports car is advertised to be able to stop in a distance of 44 m from a speed of 95 km/h. What is its [average] acceleration in m/s2? How many g's is this (g = 9.80 m/s2)?

I have 7.9 m/s2, as my answer for average acceleration, but webassign isn't accepting my answer. Can someone help me out?

 

[arildno]

How did you arrive at that answer?

 

[slayerdeus]

I used v^2-vo^2=2a(x-xo

Have to convert to m/s so vo=26.38 m/s, vf=0

 

[arildno]

Looks right to me; however:
Did you remember the sign on the acceleration?

That's about what I can think of, unless you were supposed to enter the number of g's the acceleration is in

 

[arildno]

Not 26.39?

 

[slayerdeus]

yea its 26.38888888888 m/s, no it asks for g's in another questions and when I do 7.9/9.8 it says that answer is correct. I wonder if this method is average acceleration or is it instantaneous?

 

[arildno]

It absolutely cannot be the inst. acceleration they're after; in order to know this, you would need a lot more information (effect of brakes, frictional coefficient etc.)

Since your second answer is correct, I can see only 2 possibilities:
a) The machine is wrong
b) It requires a sign for the acceleration

On second thought: Have you tried using velocity 26.4 and rounding off to one significant decimal?

 

[cookiemonster]

Still average. You have no knowledge of the interior, only the endpoints.

cookiemonster

 

[slayerdeus]

Okay, thanks mucho. It was the sign