A driver in a 1000 kg car traveling at 20 m/s slams on the brakes and skids to a stop. If the coefficient of friction between the tires and the horizontal road is 0.80, how long will the skid marks be?
A) 26 m
B) 21 m
C) 33 m
D) 24 m

Question

Quizplus · Accepted Answer

Explanation:The force of friction is given by:$$f = \mu * n$$where:* f is the force of friction* μ is the coefficient of friction* n is the normal forceThe normal force is equal to the weight of the car:$$n = mg$$where:* m is the mass of the car* g is the acceleration due to gravityThe force of friction is also equal to the force that is causing the car to accelerate:$$f = ma$$where:* m is the mass of the car* a is the acceleration of the carCombining these equations, we get:$$\mu * mg = ma$$Solving for a, we get:$$a = \mu * g$$The acceleration of the car is constant, so we can use the following equation to find the distance that the car skids:$$d = \frac{1}{2} * a * t^2$$where:* d is the distance that the car skids* a is the acceleration of the car* t is the time that the car skidsSubstituting the equation for a that we found earlier, we get:$$d = \frac{1}{2} * \mu * g * t^2$$Plugging in the given values, we get:$$d = \frac{1}{2} * 0.80 * 9.81 * t^2$$$$d = 3.92 * t^2$$To find the time that the car skids, we need to find the time it takes for the car to come to a stop. We can use the following equation:$$v = u + at$$where:* v is the final velocity of the car (0 m/s)* u is the initial velocity of the car (20 m/s)* a is the acceleration of the car (-3.92 m/s^2)* t is the time that the car skidsSolving for t, we get:$$t = \frac{v - u}{a}$$Plugging in the given values, we get:$$t = \frac{0 - 20}{-3.92}$$$$t = 5.10 s$$Substituting this value of t back into the equation for d, we get:$$d = 3.92 * (5.10)^2$$$$d = 102.6 m$$Therefore, the skid marks will be 102.6 m long.

A Driver in a 1000 Kg Car Traveling at 20