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Chapter 8: Problem 6

You are skiing down a planar skislope with an inclination \(\alpha\) with the horizontal. Your acceleration down along the slope is \(a=g \sin \alpha\). You start from a height \(h\). (a) Find your speed, \(v(t)\), measured along the slope as a function of time, \(t\). (b) Find your position, \(s(t)\), along the slope as a function of time, \(t\). (c) Find your position, \(\mathbf{r}(t)\), relative to the point you started at. (d) How long time does it take until you reach the ground at \(y=0\) ?

Short Answer

Expert verified
Part (a): \ v(t) = g \sin(\alpha) \cdot t; Part (b): \ s(t) = \frac{1}{2} g \sin(\alpha) \cdot t^2; Part (c): \ \mathbf{r}(t) = \left( \frac{1}{2} g \sin(\alpha) t^2 \cos(\alpha), \frac{1}{2} g \sin(\alpha) t^2 \sin(\alpha) \right); Part (d): \ t = \sqrt{2h / (g \sin^2(\alpha))}.

Step by step solution

01

- Determine the acceleration along the slope

Given that the acceleration along the slope \( a \) is \( a = g \sin(\alpha) \), where \(g \) is the acceleration due to gravity and \( \alpha \) is the inclination angle. This is our constant acceleration down the slope.
02

- Express velocity as a function of time

Using the equation of motion for constant acceleration, \( v(t) = u + at \). Since you start from rest (initial velocity \( u = 0 \)), it simplifies to \( v(t) = g \sin(\alpha) \cdot t \).
03

- Calculate the position along the slope

Again, using the equations of motion with constant acceleration, the position along the slope \( s(t) \) is given by \( s(t) = ut + \frac{1}{2} a t^2 \). Since \( u = 0 \), it simplifies to \( s(t) = \frac{1}{2} g \sin(\alpha) \cdot t^2 \).
04

- Find the relative position \( \mathbf{r}(t) \)

Relative to the starting point, you'll decompose the position along the slope into horizontal and vertical components. \( \mathbf{r}(t) = (s(t) \cos(\alpha), s(t) \sin(\alpha)) \). Substituting \( s(t) = \frac{1}{2} g \sin(\alpha) t^2 \, \mathbf{r}(t) = \left( \frac{1}{2} g \sin(\alpha) t^2 \cos(\alpha), \frac{1}{2} g \sin(\alpha) t^2 \sin(\alpha) \right) \).
05

- Determine time to reach the ground

Knowing the skier starts from height \( h \, and the vertical displacement is measured by \Delta y = h - \frac{1}{2} g \sin(\alpha) t^2 \sin(\alpha) \), set this equal to zero and solve for \ t. \h = \frac{1}{2} g \sin^2(\alpha)t^2 \, t^2 = \frac{2h}{g \sin^2(\alpha)}, t = \sqrt{\frac{2h}{g} \cdot \frac{1}{\sin^2(\alpha)}} \. So the time taken is \ t = \sqrt{2h / (g \sin^2(\alpha))}.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inclination Angle Analysis
Understanding the angle of inclination is crucial. The greater the angle \(\alpha\), the stronger the component of gravitational force acting along the slope. This increased force results in greater acceleration and faster velocity. The formulas for velocity and position becomes more influenced by this angle:
\r
  • Velocity: \(v(t) = g \sin(\alpha) t\)
  • Position: \[s(t) = \frac{1}{2} g \sin(\alpha) t^2\]
  • Time to reach ground: \[t = \sqrt{\frac{2h}{g \sin^2(\alpha)}}\]

So, when analyzing or solving kinematics problems, always consider how the inclination angle alters the variables. Understanding this helps with comprehending the skier’s speed and distance traveled over time.

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Most popular questions from this chapter

High speed model cars are often run in circular paths by attaching them to a wire. Here we address a car attached to a steel wire of length \(8 \mathrm{~m}\). The car starts from rest and accelerates with a tangential acceleration \(a_{t}=0.5 \mathrm{~m} / \mathrm{s}^{2}\). (a) Find the speed of the car as a function of time. (b) Find the radial acceleration of the car as function of time. (c) At what speed is the radial acceleration 100 times larger than the tangential acceleration?

A small monkey is climbing far out on a branch when it suddenly breaks. The branch does not snap off, but start to rotate about the point where it is broken. The monkey clings to the branch. What is the direction of the acceleration of the monkey?

A bead is inserted onto a thin line with an inclination \(\alpha\) with the vertical. When the bead is released, its acceleration along the line is \(a=g \cos \alpha\). (a) Find the speed, \(v(t)\), of the bead as a function of time. (b) Find the position \(s(t)\) of the bead along the line as a function of time, \(t\). (c) Find the height, \(h(t)\) of the bead as a function of time, \(t\).

During a \(200 \mathrm{~m}\) race, a sprinter is running with a speed of \(10 \mathrm{~m} / \mathrm{s}\) through the first curve. The length of the curve is \(100 \mathrm{~m}\). (a) Find the radius, \(R\), of the curve (it is a perfect half-circle). (b) Find the magnitude and direction of the acceleration \(a\) of the sprinter.

If your motion is restricted to be along a flat plane, may your acceleration be out of the plane? Explain. If your motion is restricted to be on a surface, is your acceleration restricted to be along the surface?
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