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Chapter 4: Problem 18

A moderate wind accelerates a pebble over a horizontal \(x y\) plane with a constant acceleration \(\vec{a}=\left(5.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(7.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\) . At time \(t=0,\) the velocity is \((4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}} .\) What are the (a) magnitude and (b) angle of its velocity when it has been displaced by \(12.0 \mathrm{~m}\) parallel to the \(x\) axis?

Short Answer

Expert verified
(a) 18.22 m/s, (b) 50.76掳

Step by step solution

01

Introduction

We're asked to find the magnitude and angle of the pebble's velocity after it has been displaced by 12.0 m along the x-axis under a given constant acceleration. We start with initial conditions and use kinematic equations for analysis.
02

Calculate Final Velocity in x-direction

The x-component of velocity can be determined using the kinematic equation: \[ v_x^2 = v_{0x}^2 + 2a_x s_x \] where \( v_{0x} = 4.00 \, \mathrm{m/s} \), \( a_x = 5.00 \, \mathrm{m/s}^2 \), and \( s_x = 12.0 \, \mathrm{m} \). Solving, we have: \[ v_x^2 = (4.00)^2 + 2 \times 5.00 \times 12.0 \] \[ v_x^2 = 16.00 + 120.0 = 136.0 \] \[ v_x = \sqrt{136.0} = 11.66 \, \mathrm{m/s} \].
03

Determine y-component of Velocity

To find the y-component of the velocity, use the linear velocity equation based on time: \[ v_{y} = v_{0y} + a_{y} t \] We need to find \( t \). From the x-direction equation \( x = v_{0x} \cdot t + \frac{1}{2} a_x \cdot t^2 \), we have: \[ 12.0 = 4.00 \cdot t + \frac{1}{2} \cdot 5.00 \cdot t^2 \] Rearrange and solve: \[ 2.5t^2 + 4.00t - 12.0 = 0 \] Solving gives: \[ t = 2.00 \, \mathrm{s} \]. Now calculate \( v_y \): \[ v_y = 0 + 7.00 \cdot 2.00 = 14.0 \, \mathrm{m/s} \].
04

Calculate Magnitude of the Velocity

The magnitude of the velocity vector \( \vec{v} \) can be found using Pythagoras' theorem: \[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{(11.66)^2 + (14.0)^2} \] \[ v = \sqrt{136.0 + 196.0} = \sqrt{332.0} \approx 18.22 \, \mathrm{m/s} \].
05

Find Angle of Velocity with x-axis

The angle \( \theta \) can be found using the tangent function: \[ \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{14.0}{11.66}\right) \] \[ \theta \approx 50.76^\circ \].

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Key Concepts

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

Projectile motion
Projectile motion is a form of motion where an object moves along a curved path under the influence of gravity and other forces such as wind. In this specific problem, the pebble moves in a horizontal plane due to a moderate wind accelerating it, providing a unique angle on projectile motion. The pebble experiences an initial velocity and constant acceleration over time, resulting in a displacement along the x-axis. To fully understand projectile motion, consider how the pebble's movement is broken into two components: horizontal and vertical, each affected by different factors. These movements combined create the overall path of the pebble.
Velocity
Velocity is a vector quantity, meaning it has both magnitude and direction. In the given problem, the pebble starts with an initial velocity in the horizontal (x) direction. As time progresses, both its x and y components of velocity change due to continuous acceleration. By knowing the initial conditions and the acceleration, we can calculate both components of its final velocity using the kinematics equations. In our example, these equations enable the calculation of the pebble's velocity at a specific displacement, showing how velocity develops under constant acceleration.
Acceleration
Acceleration is the rate at which an object changes its velocity. It is a vector quantity with both magnitude and direction. In the pebble example, the constant acceleration is provided by the wind, with values of 5.00 m/s虏 in the x-direction and 7.00 m/s虏 in the y-direction. Under these conditions, acceleration influences how quickly the pebble speeds up or slows down in each direction. Understanding acceleration allows for the calculation of how long it takes for the pebble to reach a specific velocity, which in turn helps us determine other properties of its motion.
Pythagorean theorem
The Pythagorean theorem is crucial in calculating the magnitude of the velocity vector in our problem. This theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. In the context of our exercise, the horizontal and vertical components of the pebble's velocity form the sides of a right triangle, with the overall velocity being the hypotenuse. By applying the Pythagorean theorem, we can find the magnitude of the velocity based on its x and y components, which is a critical step in understanding the pebble's resulting speed.
Trigonometry
Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. It plays a significant role in determining the angle of the velocity vector with respect to the x-axis in our pebble exercise. By using the tangent function, which relates the opposite and adjacent sides of a right triangle, we can calculate the angle of the pebble's velocity. This angle provides insight into the direction of the pebble's motion and is important when analyzing its trajectory. Understanding trigonometry is essential for solving problems involving angled motion and vector components.

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Most popular questions from this chapter

When a large star becomes a supernova, its core may be compressed so tightly that it becomes a neutron star, with a radius of about \(20 \mathrm{~km}\) (about the size of the San Francisco area). If a neutron star rotates once every second, (a) what is the speed of a particle on the star's equator and (b) what is the magnitude of the particle's centripetal acceleration? (c) If the neutron star rotates faster, do the answers to (a) and (b) increase, decrease, or remain the same?

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Long flights at midlatitudes in the Northern Hemisphere encounter the jet stream, an eastward airflow that can affect a plane's speed relative to Earth's surface. If a pilot maintains a certain speed relative to the air (the plane's airspeed), the speed relative to the surface (the plane's ground speed) is more when the flight is in the direction of the jet stream and less when the flight is opposite the jet stream. Suppose a round-trip flight is scheduled between two cities separated by \(4000 \mathrm{~km}\), with the outgoing flight in the direction of the jet stream and the return flight opposite it. The airline computer advises an airspeed of \(1000 \mathrm{~km} / \mathrm{h},\) for which the difference in flight times for the outgoing and return flights is \(70.0 \mathrm{~min} .\) What jet-stream speed is the computer using?
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