Expert: James J. Kovalcin - 9/14/2005

Question
A race car can be slowed with a constant acceleration of -11 m/s(squared).  If the car is going 55 m/s, how many meters will it take to stop?

Answer
ALL questions of this type are solved in exactly te same way.
There are two kinematics equations:
Df=1/2*a*t^2+Vo*t+Do  and  Vf=a*t+Vo
Which contain exactly 6 variables:
Do, Df, Vo, Vf, a & t.
If you can identify 4 of these variables you can always solve for the other 2. In this case the known quantities are:
Do=0m [Assume this is zero unless the question indicates otherwise], Vo=55m/s [given], a=-11m/s^2 [given] and Vf=0 [the car stops at the end].
This leaves you with 2 unknowns: Df=? and t=?.
Substituting the known values into the velocity equation:
Vf=a*t+Vo becomes 0=-11*t+55 solve for t!
Now use the displacement equation:
Df=1/2*a*t^2+Vo*t+Do becomes Df=-11*t^2+55*t+0band solve for Df.
ALL kinematics problems can be done using the same approach! Really!