A car of mass 1000.0 kilograms is traveling along a level road at 100.0 kilometers per hour when its brakes are applied. Calculate the stopping distance if the coefficient of kinetic friction of the tires is 0.500. Neglect air resistance.
Let’s begin by highlighting some of the important information we’ve been given. We’re told that the mass of the car is 1000.0 kilograms, we’ll call that 𝑚, and that it’s moving at a speed of 100.0 kilometers per hour. We’ll call that speed 𝑣 sub 𝑖. We’re told that the coefficient of kinetic friction between the tires and the road is 0.500. We’ll call that 𝜇 sub 𝑘. And overall, we want to solve for the distance it will take the car to come to a complete stop. Let’s call that distance 𝑑.
We can draw a diagram of our scenario. Our car of mass 𝑚 is driving along a level road at the speed 𝑣 sub 𝑖 when it hits the brakes. We can assume this means the wheels stop rotating and begin to slide across the road surface. It’s at that point that a frictional force begins to oppose the forward motion of the car. This force causes the car to decelerate and eventually come to a stop a distance 𝑑 away from where the brakes were initially applied.
As we begin to solve for that distance, let’s recall the equation for frictional force. The friction force acting on an object is equal to the normal force on that object times the coefficient of friction, 𝜇. Let’s apply this equation to our particular scenario. The friction force on the car equals the normal force on the car multiplied by 𝜇 sub 𝑘, the coefficient of kinetic friction. We’re told the car is on a level surface, and we know it’s not accelerating vertically, which means that the weight force of the car, its mass times gravity, is equal and opposite to the normal force acting on the car.
Because we are considering only force magnitudes in this equation for friction force, this means we can substitute the car’s mass times the acceleration due to gravity in for 𝐹 sub 𝑁. Remember that this equation applies to a force that’s in the horizontal direction. And because we’re neglecting air resistance, the frictional force is the only horizontal force acting on the car. Let’s recall what Newton’s second law says about net forces acting on an object. This law tells us that the net force on an object is equal to its mass times its acceleration. In other words, the frictional force on the car is equal to the car’s mass times its horizontal acceleration.
We’re interested in solving for that acceleration because that will help us figure out how long it takes the car to come to a stop. Since the mass of the car is common to the right and middle terms of our equation, we can cancel it out. So we see that the car’s acceleration in the horizontal direction is equal to 𝑔, the acceleration due to gravity, times 𝜇 sub 𝑘, the coefficient of kinetic friction. We assume that 𝑔 is equal to exactly 9.8 meters per second squared, and 𝜇 sub 𝑘 is given as 0.500. Multiplying these values together, we find that the car’s horizontal acceleration is 4.9 meters per second squared.
Over the course of the car coming to a stop, this acceleration it experiences doing the friction force is constant. That means that the kinematic equations of motion apply to this scenario. Let’s recall those four equations. These equations of motion describe scenarios where acceleration is constant. As we look for an equation that matches what we’ve been given and what we’re looking for, we see that the second equation from the top is a good match. When we apply it to our scenario, we see that 𝑣 𝑓, the final speed of the car, is zero, so 𝑣 𝑓 squared goes to zero. 𝑣 zero, the car’s initial speed, we know as 𝑣 sub 𝑖.
We’ve solved for 𝑎, the car’s acceleration, which is 4.9 meters per second squared. And 𝑑 is the distance the car travels before it comes to a stop that we’re solving for. Going back to our diagram, let’s define motion in the car’s original direction of motion as positive. That means that because the car is accelerating in the opposite direction, its acceleration is not positive 4.9 meters per second squared, but negative 4.9 meters per second squared. We return to our kinematic equation where we want to solve for 𝑑, the distance the car travels before it comes to a stop.
Let’s subtract 𝑣 sub 𝑖 squared from both sides, which cancels this term from the right-hand side of the equation. We then divide both sides by two times the acceleration 𝑎, which cancels those two factors from the right-hand side of our equation. We’re left with the result that 𝑑 is equal to negative 𝑣 sub 𝑖 squared divided by two times 𝑎. We’ve solved for 𝑎 and can insert that into this equation. But before we put in 𝑣 sub 𝑖, we’ll want to convert the units of 𝑣 sub 𝑖 into meters per second. To convert 100.0 kilometers per hour into meters per second, we can first multiply by 1000 meters per one kilometer.
Doing this cancels out the units of kilometers. We can then multiply by one hour per 3600 seconds. Doing this cancels out the units of hours. We’re left with a speed in units of meters per second. Multiplying these fractions through, that speed is 27.78 meters per second. Now we’re ready to substitute in for 𝑣 sub 𝑖 and 𝑎 in the equation for 𝑑. When we substitute in those values and then enter these terms on our calculator, we find a value for 𝑑 of 78.7 meters. To three significant figures, that’s how far the car would skid before it comes to a complete stop.