Question

A car is traveling at 43.3 mi/h on a horizontal highway. The acceleration of gravity is 9.8 m/s 2 . If the coefficient of friction between road and tires on a rainy day is 0.097, what is the minimum distance in which the car will stop? (1 mi = 1.609) Answer in units of m. Please help!

Answer 1

The key things to recognize here is that the car is moving with a constant velocity, v = 43.3 mi/h, meaning that the Net Force is 0 in the X and Y directions. Now we have \[F _{xNet} = m*a _{x} = -F _{friction} = -F _{normal}*\mu _{k}\], where the normal force comes from the net forces in the Y-direction. In this case \[F _{N} = -F _{g} = -m*g\] you place this Force Normal in to the above equation and solve for the acceleration, the mass will cancel out so don't worry about that. Now with the acceleration know we look to kinematics to solve for the distance. Using \[V ^{2} = v _{0}^{2} -2*a _{x}*\Delta x\] where delta x is what you are solving for all you have to do is convert 43.3 mi/h into m/s and solve, remembering that V is zero. Your final equation will be \[\Delta x = v _{0}^{2}/(2*a _{x})\].

Answer 2

Thanks so much

Answer 3

You are very welcome.