A skier is pulled up a hill with an inclination \(\alpha\) with the horizontal. He is pulled with a constant acceleration of \(a=2 \mathrm{~m} / \mathrm{s}^{2}\) along the hill and starts from rest at the bottom of the hill. (a) Find the speed, \(v(t)\), of the skier measured along the slope as a function of time, \(t\). (b) Find the position, \(s(t)\), of the skier measured as a distance from the starting point after a time \(t\). (c) Find the position, \(\mathbf{r}(t)\), of the skier in the \(x y\)-coordinate system, where \(x\) is the horizontal axis and \(y\) is the vertical axis. (d) Use the vector position, \(\mathbf{r}(t)\), to find the speed of the skier, and compare with the results you found above.

Constant acceleration, also known as uniform acceleration, is a type of acceleration where the rate of change of velocity remains consistent over time. In our skier problem, the skier is pulled up the hill with an acceleration of \(a = 2 \mathrm{~m/s^2}\). This means that for every second, the skier’s velocity increases by 2 meters per second. When an object such as the skier undergoes constant acceleration, we can use a simple formula for velocity:

• \( v(t) = v_0 + at \)

Here, \(v_0\) is the initial velocity, which is zero since the skier starts from rest, hence the velocity function simplifies to \(v(t) = 2t\). This means after 1 second, the speed is 2 m/s, after 2 seconds, it’s 4 m/s, and so on. This linear relationship makes calculations straightforward and predictable.

The velocity function describes how speed changes with time under the influence of constant acceleration. For our skier, who starts from rest, the velocity function can be derived from the equation: \(v(t) = v_0 + at\). With \(v_0 = 0\) and \(a = 2 \mathrm{~m/s^2}\), we get:

• \(v(t) = 2t \)
  

This means the skier’s speed only depends on time spent accelerating. As time progresses, velocity increases linearly. Keep in mind, velocity is a vector quantity having both magnitude and direction. On the slope, this direction is along the incline. This function is critical for predicting how quickly the skier moves up the hill at any given moment.

The position function tells us the skier’s distance from the starting point as time progresses. With constant acceleration, the position is found using the formula:

• \(s(t) = s_0 + v_0t + \frac{1}{2}at^2 \)

Given that \(s_0 = 0\) and \(v_0 = 0\), it simplifies to:

• \(s(t) = \frac{1}{2}at^2 \)

In this problem, substituting \(a = 2 \mathrm{~m/s^2}\) into the equation, we get:

• \(s(t) = t^2 \)
  

This reveals that the skier's position grows quadratically with time, meaning it speeds up more quickly as time goes on. This formula helps us find out how far the skier has traveled up the slope at any given time.

Knowing the skier’s position along the slope, we can find its coordinates in a standard \(xy\) coordinate system using vectors. Using trigonometry, the horizontal \(x\) and vertical \(y\) components for the position are derived from the total displacement along the plane. If the skier’s position along the slope is \(s(t) = t^2\), then:

• \(x(t) = t^2 \cos\alpha\)
  
• \(y(t) = t^2 \sin\alpha\)
  

These components form the vector position \(\mathbf{r}(t)\) which is \(\mathbf{r}(t) = (t^2 \cos\alpha \mathbf{i}, t^2 \sin\alpha \mathbf{j})\).
Here, \(\alpha\) is the angle of the incline. This vector represents the skier’s position in two dimensions and is essential for understanding movements relative to both horizontal and vertical directions.

Trigonometry plays a crucial role in physics problems like this one. When dealing with inclined planes, knowing the angle of incline, \(\alpha\), allows us to decompose movements into horizontal and vertical components. For instance, the skier’s displacement along the slope \(s(t)\) can be split into:

• Horizontal component: \(x(t) = s(t) \cos\alpha \)
• Vertical component: \(y(t) = s(t) \sin\alpha \)
  

Through these equations, we understand how much of the skier’s movement goes horizontally versus vertically. It's essentially applying basic trigonometric identities where \(\cos\alpha\) gives the ratio of adjacent to hypotenuse (horizontal part), and \(\sin\alpha\) gives the ratio of opposite to hypotenuse (vertical part). It’s a fundamental skill in physics to break down vectors into components to analyze movements and forces more conveniently.