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Chapter 3: Problem 8

A skateboarder, starting from rest, rolls down a 12.0 -m ramp. When she arrives at the bottom of the ramp her speed is \(7.70 \mathrm{~m} / \mathrm{s}\). (a) Determine the magnitude of her acceleration, assumed to be constant, (b) If the ramp is inclined at \(25.0^{\circ}\) with respect to the ground, what is the component of her acceleration that is parallel to the ground?

Short Answer

Expert verified
(a) The acceleration is \(2.47 \ \mathrm{m/s^2}\). (b) The acceleration parallel to the ground is \(2.24 \ \mathrm{m/s^2}\).

Step by step solution

01

Identifying Known Variables

First, we identify the given information and assign variables. The initial velocity \( v_i \) is \( 0 \ \mathrm{m/s} \) (since she starts from rest), the final velocity \( v_f \) is \( 7.70 \ \mathrm{m/s} \), the displacement \( s \) down the ramp is \( 12.0 \ \mathrm{m} \), and the angle of inclination \( \theta \) is \( 25.0^{\circ} \).
02

Calculate Acceleration Using Kinematics

To find the acceleration \( a \), we use the kinematic equation: \( v_f^2 = v_i^2 + 2as \). Substituting the known values: \( 7.70^2 = 0 + 2a \times 12.0 \). Solve for \( a \).
03

Simplifying the Equation

The equation simplifies to \( 59.29 = 24a \). Solving for \( a \) gives \( a = \frac{59.29}{24} \approx 2.47 \ \mathrm{m/s^2} \).
04

Decompose Acceleration into Parallel Components

The component of acceleration parallel to the ground \( a_{x} \) is given by \( a_{x} = a \cos \theta \). Here, \( a = 2.47 \ \mathrm{m/s^2} \) and \( \theta = 25.0^{\circ} \).
05

Calculate the Parallel Component

Substitute the known values into the equation: \( a_{x} = 2.47 \cos(25^{\circ}) \). Calculate to find \( a_{x} \approx 2.24 \ \mathrm{m/s^2} \).

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Key Concepts

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

Acceleration
Acceleration is a key concept in kinematics and is defined as the rate at which an object changes its velocity. In the example of the skateboarder, we start by knowing her initial and final velocities and the ramp's length. Since she starts from rest, her initial speed is 0 m/s, and she accelerates to a final speed of 7.70 m/s by the end of the ramp's 12.0 m length. Here, acceleration can be found using kinematic equations that relate these values. It is crucial to remember that acceleration can either speed up or slow down an object depending on the direction of the applied force relative to the motion.
Inclined Plane
An inclined plane is a flat surface tilted at an angle to the horizontal. In physics, this incline makes it easier to analyze forces and motion in two dimensions, rather than along the slope.

The skateboarder rolls down an inclined plane of 25 degrees. This angle significantly affects how gravity contributes to her motion. Gravity pulls directly downward, but the incline means this force is distributed into components parallel and perpendicular to the ramp. The steeper the incline, the more pronounced the acceleration effect along the ramp due to gravity.
Constant Acceleration
Constant acceleration means that the object's velocity changes at a uniform rate throughout its motion. In this context, when calculating the skateboarder's motion down the ramp, we assume the acceleration remained constant.

This assumption simplifies the maths because certain kinematic equations only hold under constant acceleration conditions, such as the one used here: \[ v_f^2 = v_i^2 + 2as \]Using this formula, we calculated her acceleration to be approximately 2.47 m/s虏. Assuming constant acceleration allows us to predict the skateboarder's velocity and position at any point on the ramp.
Kinematic Equations
Kinematic equations are indispensable tools in classical mechanics, helping to describe motion with uniform acceleration. They relate quantities like displacement, initial velocity, final velocity, time, and acceleration.

In the problem of the skateboarder, one such equation \[ v_f^2 = v_i^2 + 2as \]is employed to calculate the acceleration with known final velocity, initial velocity, and displacement.
  • They enable precise predictions of future motion based on current and past states.
  • They require assumptions like constant acceleration for effectiveness.

Understanding these equations helps to solve complex motion problems efficiently, revealing much about the dynamics at work.

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Most popular questions from this chapter

A major-league pitcher can throw a baseball in excess of \(41.0 \mathrm{~m} / \mathrm{s}\). If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches a catcher who is \(17.0 \mathrm{~m}\) away from the point of release?

(a) When a projectile is launched horizontally from a rooftop at a speed \(v_{0 x}\), does its horizontal velocity component ever change in the absence of air resistance? (b) Can you calculate the horizontal distance \(D\) traveled after launch simply as \(D=v_{0 x} t\), where \(t\) is the fall time of the projectile? (c) In calculating the fall time, is the vertical part of the motion just like that of a ball dropped from rest? Explain each of your answers. A criminal is escaping across a rooftop and runs off the roof horizontally at a speed of \(5.3 \mathrm{~m} / \mathrm{s}\), hoping to land on the roof of an adjacent building. The horizontal distance between the two buildings is \(D,\) and the roof of the adjacent building is \(2.0 \mathrm{~m}\) below the jumping- off point. Find the maximum value for \(D\).

A volleyball is spiked so that it has an initial velocity of \(15 \mathrm{~m} / \mathrm{s}\) directed downward at an angle of \(55^{\circ}\) below the horizontal. What is the horizontal component of the ball's velocity when the opposing player fields the ball?

A jetliner can fly 6.00 hours on a full load of fuel. Without any wind it flies at a speed of \(2.40 \times 10^{2} \mathrm{~m} / \mathrm{s} .\) The plane is to make a round-trip by heading due west for a certain distance, turning around, and then heading due east for the return trip. During the entire flight, however, the plane encounters a \(57.8-\mathrm{m} / \mathrm{s}\) wind from the jet stream, which blows from west to east. What is the maximum distance that the plane can travel due west and just be able to return home?

An Olympic long jumper leaves the ground at an angle of \(23^{\circ}\) and travels through the air for a horizontal distance of \(8.7 \mathrm{~m}\) before landing. What is the takeoff speed of the jumper?
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