### Video Transcript

A roller coaster car starts to move from rest at the top of a track that is 30.0 meters long and inclined at an angle 20.0 degrees below the horizontal. Assume that friction is negligible. What is the magnitude of the acceleration of the car? How much time passes before the car reaches the bottom of the track?

Let’s start by highlighting some of the important information given. We’re told that the track is 30.0 meters long; we’ll call that 𝐿. We’re also told that the track is inclined 20.0 degrees below the horizontal; we’ll call that angle 𝜃. In part one, we want to solve for the magnitude of the car’s acceleration; we’ll call that 𝑎. And in part two, we want to find how much time the car takes to reach the bottom of the track; we’ll call that 𝑡.

Along with assuming that friction is negligible, we’ll also assume that the acceleration due to gravity, 𝑔, is exactly 9.8 meters per second squared. Let’s start out by drawing a diagram of the car on the track. Here we see the roller coaster car on the track, with positive motion directions defined as normal to the track that is perpendicular and down the track parallel to it. As we work towards solving for the car;s acceleration, let’s recall Newton’s second law.

The second law states that the net force acting on an object is equal to its mass times its acceleration. To solve for acceleration, we’ll first look at the forces acting on the car. If we draw a free body diagram of the car, the only two forces acting on the car when friction is negligible are the force of gravity, 𝐹 sub 𝑔, and the normal, Force 𝐹 sub 𝑁. If we break the gravitational force into components that are parallel and perpendicular to the incline, then we create a right triangle where the hypotenuse of the triangle is the force of gravity and the angle between the hypotenuse and the adjacent side is equal to 𝜃, the given value of 20.0 degrees.

If we call motion that’s parallel to the track motion in the 𝑥-direction and motion perpendicular to the track motion in the 𝑦-direction, then we can label the components of the gravitational force 𝐹 sub 𝑥 and 𝐹 sub 𝑦. From the diagram, we can see that it’s the 𝑥-component of the gravitational force that pushes the car down the track. That force component is equal to the mass of the car times 𝑔 multiplied by the sine of 𝜃. And because that is the only force acting in the 𝑥-direction, by Newton’s second law, it’s equal to the mass of the car times its acceleration.

So the mass of the car cancels out, and we see that the car’s acceleration is equal to 𝑔 times the sine of 𝜃. Plugging in 9.8 meters per second squared for 𝑔 and 20.0 degrees for 𝜃, we find that the car’s acceleration is equal to 3.35 meters per second squared. That’s the acceleration of the car down the track. Next, we wanna solve for the time it takes the car to move down the track from the top to the bottom, knowing that the track is a length 𝐿 of 30.0 meters long.

Since the car accelerates uniformly down the track, we can apply the kinematic equations to the motion of the car. Of the four kinematic equations, the third one, that displacement is equal to initial speed times time plus one-half acceleration times time squared, applies to our situation. In our case, 𝑑 is equal to 𝐿, the length of the track. The car starts from rest, so it’s initial speed, 𝑣 zero, is equal to zero, and that term cancels out.

We’ve solved for the acceleration 𝑎 in the previous step, and 𝑡 is what we are solving for now. So 𝐿 equals one-half 𝑎𝑡 squared. If we multiply both sides by two divided by 𝑎, then acceleration and the factor of two cancel out of the right side of our equation. And if we then take the square root of both sides, that square root cancels with the square term on the right, leaving us with an equation for 𝑡, that 𝑡 is equal to the square root of two times 𝐿 divided by 𝑎.

Plugging in our values for 𝐿 and 𝑎, when we enter these numbers on our calculator, we find a time value of 4.23 seconds. That’s how long it takes the car to move down this section of track starting from rest.