A car has a mass of 1500kg. If the driver applies the breaks, the maximum amount of friction force that can be applies without skidding is 7000N. If the car is traveling at 25m/s, what is the shortest distance the car can stop safely?
I'm assuming it has something to do with F=mα and using that answer to solve for distance using the kinematics equation: Vfinal(squared)=Vinitial+2α[(X-X[initial]
The Attempt at a Solution
I divided 7000N by 1500kg to cancel out the kg, to get 4.67 m/s(squared?) I assumed that was what my acceleration and said my final velocity should be 0. So my equation worked out to be 0=25+2×4.67(X-0). then I tried to solve for x and I got 2.67m, then I transposed 25 for acceleration and 4.67 for initial velocity and got .0467m which is way too small, the correct answer is 67m (according to masteringphysics). Even after getting the solution and plugging that into the equation for X ; the solution doesn't equal 0 like it should. I tried to solve using the position kinematics equation X=Xinitial+V(initial)×T(squared)+.5α×T(squared) but I couldn't make it work, since I wasn't given a time interval to work with.