### Video Transcript

A car of mass π tons was initially at rest on a hill inclined at an angle π to the horizontal, where sine of π equals one-half. After 100 seconds, its velocity was 21 metres per second. Calculate the resistance per ton of the carβs mass. Take π to equal 9.8 metres per second squared.

We can call the elapsed time during which the car is speeding up 100 seconds π‘. And the speed it achieves after that time 21 metres per second weβll call π£ sub π. Weβre told that the sine of the angle of inclination of the hill on which the car rests is one-half. We want to calculate the resistance per ton of the carβs mass. Weβll call that value πΉ sub π over π.

Letβs begin our solution by drawing a sketch of this scenario. Weβre told that we begin with a car at rest on a hillside inclined at an angle π from the horizontal. At π‘ equals zero, the car is released and begins to slide downhill. After a time of 100 seconds, the car has achieved a particular speed 21 metres per second downhill. Knowing this about the carβs descent, we want to solve for the resistive force β weβve called it πΉ sub π β applied to the car per ton of its mass π.

To start on our solution, letβs draw in the forces that are acting on the car while it descends. First, there is the force of gravity acting down. Thatβs equal to the mass of the car times the acceleration due to gravity π, which weβre told to treat as exactly 9.8 metres per second squared. There is also a normal force pushing the car perpendicularly away from the plane of the hill. We call that πΉ sub π. And finally, there is a resistive force due to friction, which weβve called πΉ sub π, acting on the car and opposing its downward motion.

To solve for that force per ton of the carβs mass, letβs begin by recalling Newtonβs second law of motion. The second law says that the net force acting on an object is equal to that objectβs mass multiplied by its acceleration. If we place a pair of coordinate axes on our diagram so that π₯ is acting up the hill and π¦ is positive perpendicularly to the plane, then we become able to apply the second law to both directions.

Letβs consider the forces in the π₯-direction. Two of our three forces in our diagram act along this axis πΉ sub π and a component of the weight force ππ. If we divide our weight force ππ into π¦- and π₯-components, then the right triangle formed has an angle of π in its upper corner. This means when we apply Newtonβs second law to forces in the π₯-direction, we can write πΉ sub π minus ππ times the sine of π is equal to π times π.

If we then divide both sides of our equation by the mass of the car π, that term cancels out of two of our three terms and we now have πΉ sub π over π in the far left, which is what we want to solve for. We can rearrange our equation. So it reads πΉ sub π over π is equal to π sine π plus π.

We know the gravitational acceleration π and we also know the sine of π. But what about the acceleration π? We can recall that acceleration is defined as a change in velocity over change in time. Weβre told that our car starts out from rest. So our initial velocity is zero and that it attains a final velocity π£ sub π of 21 metres per second. All this happens over the course of a time of 100 seconds.

So when we plug in π£ sub π and π‘, we see that π is negative 21 metres per second divided by 100 seconds or negative 0.21 metres per second squared. Weβre now ready to plug this value in for π. And we know π and the sine of π.

When we enter these values on our calculator, we find that πΉ sub π over π is equal to 4.69 newtons per ton. Thatβs the resistive force of the hill on the car per ton of the carβs mass.