Video Transcript
A car was moving down a hill inclined at an angle 𝜃 to the horizontal, where the sin of 𝜃 equals four over 75. When its engine was off, it moved at a constant speed. If the same car was moving up the same slope at 2.8 meters per second and its engine cut out, how far would it move before it came to rest? Assume that the magnitude of the resistance to its movement is the same while both ascending and descending. Take 𝑔 to equal 9.8 meters per second squared.
Okay, see that this is our hill inclined at an angle 𝜃 to the horizontal. And we’re told that the sin of this angle equals four seventy-fifths. There’s a car moving on this hill. And we have two bits of information about it. In the first case, we’re told that when the car’s engine is shut off and it rolls downhill, that rolling happens at a constant speed. We then imagine a different scenario where that same car on the same slope is moving uphill at 2.8 meters per second. If its engine cuts off so that the car is no longer powering its motion, we want to know how far up the hillside, and we can call this distance 𝑑, the car will go before it comes to rest.
To get started on our solution, let’s clear some space to work. And we’ll begin by considering this case of the car as it moves uphill. After moving along at this given initial speed, we’re told that the car’s engine cuts off. So now it’s just rolling under its own momentum. If we were to sketch in the forces that act on the car as it still rolls uphill, we can say that it’s subject to a weight force, its mass times the acceleration due to gravity, a reaction or normal force that we’ll call 𝑅 acting perpendicularly to the plane, and lastly there’s a frictional force. Because the car is still moving uphill in this snapshot and friction always opposes the direction of motion, that force will act down the incline.
If we were to focus only on the forces acting parallel to this incline, that would include the friction force as well as this component here of the weight force. By Newton’s second law of motion, the sum of these forces is equal to our object’s mass multiplied by its acceleration in the direction of the forces. Say that we set up a coordinate system so that the positive 𝑥-direction is down the incline and the positive 𝑦-direction is perpendicularly away from it. By this sign convention, our frictional force 𝐹 has a positive value and so does the component of our weight force acting in what we’ve called the 𝑥-direction.
To solve for that component, we can realize that this angle here in the right triangle created by the components of our weight force is identical to this angle in our plane. That is, it’s equal to 𝜃. Therefore, this component in the 𝑥-direction equals 𝑚 times 𝑔 times the sin of 𝜃. Since these are the only forces acting in what we’ve called the 𝑥-direction on our car, their sum is equal to the car’s mass multiplied by its acceleration in this dimension. Now the sin of 𝜃 is given as four over 75. And along with that, we can say that in general the friction force on a moving object is equal to the coefficient of friction times the reaction force acting on that object. And from our diagram, the reaction force on the car is equal in magnitude to this component of its weight force.
That component equals 𝑚 times 𝑔 times the cos of 𝜃. So by multiplying this by 𝜇, we now have an expression for the frictional force acting on the car. This force plus 𝑚 times 𝑔 times four over 75, recall that that’s the sin of 𝜃, is equal to 𝑚 times the acceleration of our car in the 𝑥-direction. Notice that the mass of our car appears in all three of these terms. And therefore, we can effectively cancel it out. Next, we can work towards solving for what the cos of 𝜃 is equal to. With the sin of 𝜃 equaling four over 75, we can say that those are the effective side lengths of these two sides of this right triangle. And by the Pythagorean theorem, this third side equals the square root of 75 squared minus four squared or precisely the square root of 5609.
We know then that the cos of 𝜃 is equal to this value divided by the length of the hypotenuse. Substituting this in for the cos of 𝜃 and factoring out the acceleration due to gravity, which appears in both terms on the left-hand side of our expression, we arrive at this result for the acceleration of our car in the 𝑥-direction. Recall that this is as the car is slowing down, going uphill, and coming to a stop. We know the acceleration due to gravity, but we don’t know what 𝜇 is. That’s the coefficient of friction between the car’s tires and the incline.
At this point, though, we can recall a very important piece of information we were given in our problem statement. We were told that when our car was on this slope when the engine cut off and starting from rest, it would roll downhill moving at a constant speed. That is, its acceleration was zero. In this case, with our car rolling downhill, the forces acting on it would look like this. Note that the weigh force, 𝑚𝑔, and the reaction force, 𝑅, are the same as before but that now our frictional force, we’ve called it 𝐹 two, is acting up the incline, as always opposing the motion of our object. And because our car is moving at a constant speed as it rolls downhill, we can say that the 𝑥-component of the weight force and the frictional force perfectly balance one another out.
In other words, 𝑚 times 𝑔 times the sin of 𝜃, the component of the weight force pulling the car downhill, is equal to 𝜇 times 𝑚 times 𝑔 times the cos of 𝜃, the frictional force opposing this motion. Notice that both the mass of this car and the acceleration due to gravity appear on both sides of this equation and therefore effectively cancel out. If we then divide both sides of the equation by the cos of 𝜃 canceling this factor on the right, we can make use of the mathematical identity that the sine of an angle divided by the cosine of that same angle equals the tangent of the angle. Therefore, when all the dust settles, the coefficient of friction 𝜇 equals the tan of 𝜃. And from our sketch in the bottom left, we know that that’s equal to four divided by the square root of 5609.
So now we do have an expression for the coefficient of friction that we can substitute into our equation for the acceleration of our car as it moves uphill. Making this substitution, we see that the factors of the square root of 5609 in numerator and denominator cancel. And we find that the acceleration of our car in the 𝑥-direction as it moves uphill is eight seventy-fifths times the acceleration due to gravity. We’ve made great progress, but we still want to find out what is this distance that the car travels before it comes to rest. To figure this out, we can note that because our car is accelerating at a constant rate, its motion is described by what are called the equations of motion.
One of these four equations tells us that the final velocity of an object squared is equal to its original velocity squared plus two times its acceleration multiplied by its displacement. In our scenario, because our car is coming to rest at the end, we know that 𝑣 sub f equals zero. So we can write that zero equals 𝑣 sub zero squared plus two times 𝑎 times 𝑑 or that 𝑑 equals negative 𝑣 sub zero squared over two times 𝑎. Notice that there’s a minus sign leading off this result. That’s there because technically we’re calculating a displacement here rather than a distance. In order to calculate a distance instead, all we need to do is take the absolute value of this fraction.
Plugging in for 𝑣 zero and 𝑎 and leaving out their units, we get this expression. And if we recognize that 𝑔 is equal to 9.8 meters per second squared, then we can move ahead with calculating the distance 𝑑 that this car rolls before coming to a stop. Entering this expression on our calculator, 3.75 is our result. And this distance has units of meters. On this particular hill then, if our car was moving uphill at 2.8 meters per second and then its engine cut off, it would roll 3.75 meters further uphill before coming to a stop.
