Fresh features from the #1 AI-enhanced learning platform.Try it free
Fresh features from the #1 AI-enhanced learning platformCrush your year with the magic of personalized studying.Try it free
Question

# In a downhill ski race, surprisingly, little advantage is gained by getting a running start. (This is because the initial kinetic energy is small compared with the gain in gravitational potential energy on even small hills.) To demonstrate this, find the final speed and the time taken for a skier who skies 70.0 m along a $30^{\circ}$ slope neglecting friction: (a) Starting from rest. (b) Starting with an initial speed of 2.50 m/s. (c) Does the answer surprise you? Discuss why it is still advantageous to get a running start in very competitive events.

Solution

Verified
Step 1
1 of 2

Let's first find the general formula for time and then just plug in the numbers. The acceleration is due to gravity and is equal to:

$a = g \sin \theta$

This gives the distance of:

$d = v_0 t + \frac{1}{2}at^2 = v_0 t + \frac{1}{2} g t^2 \sin \theta$

Solving this quadratic equation for $t$:

$t = \frac{-v_0 + \sqrt{ v_0^2 + 2g d \sin \theta }}{ g \sin \theta }$

We can only take the solution with a "+" sign since negative time makes no sense.

a)

$t_1 = \frac{\sqrt{2 \cdot 9.81 \text{ m/s}^2 \cdot 70 \text{ m} \cdot \sin(30^\circ) }}{ 9.81 \text{ m/s}^2 \cdot \sin(30^\circ) } = \boxed{ 5.34 \text{ s}}$

b)

$t_2 = \frac{-2.5 \text{ m/s} + \sqrt{ (-2.5 \text{ m/s})^2 + 2 \cdot 9.81 \text{ m/s}^2 \cdot 70 \text{ m} \cdot \sin(30^\circ) }}{ 9.81 \text{ m/s}^2 \cdot \sin(30^\circ) } = \boxed{ 4.86 \text{ s}}$

c) The answer isn't suprising, it's explained at the beggining of the task why the difference isn't big. In competitive events the difference in time comes to one hundreds of a second so every contribution matters.

## Recommended textbook solutions

#### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

4th EditionISBN: 9780133942651 (4 more)Randall D. Knight
3,508 solutions

#### Mathematical Methods in the Physical Sciences

3rd EditionISBN: 9780471198260Mary L. Boas
3,355 solutions

#### Fundamentals of Physics

10th EditionISBN: 9781118230718 (5 more)David Halliday, Jearl Walker, Robert Resnick
8,971 solutions

#### College Physics

1st EditionISBN: 9781938168000 (4 more)Paul Peter Urone, Roger A Hinrichs
3,170 solutions