/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 A moderate wind accelerates a pe... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \((4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{i}}\). 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
Magnitude: approximately 14.70 m/s; Angle: approximately 60.3 degrees.

Step by step solution

01

Determine the time taken for x displacement

Since we know that the displacement along the x-axis is 12 m, we can use the kinematic equation for displacement:\[ s_x = v_{ix}t + \frac{1}{2} a_x t^2 \]Where \( s_x = 12.0\, \text{m} \), \( v_{ix} = 4.00\, \text{m/s} \), and \( a_x = 5.00\, \text{m/s}^2 \). Plugging in the values gives:\[ 12.0 = 4.00t + \frac{1}{2} \times 5.00 \times t^2 \]Simplifying further:\[ 12.0 = 4.00t + 2.50t^2 \]This is a quadratic equation in the form \( 2.5t^2 + 4.00t - 12 = 0 \). Solve for \( t \) using the quadratic formula:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \( a = 2.5 \), \( b = 4.00 \), \( c = -12.0 \). Calculate to find the positive value of \( t \).
02

Calculate the final velocity components

With the time \( t \) calculated, use the velocity equations for each direction:For the x-component, use:\[ v_x = v_{ix} + a_x t \]Substitute \( v_{ix} = 4.00 \text{ m/s} \), \( a_x = 5.00 \text{ m/s}^2 \), and the computed \( t \) from Step 1.For the y-component, since the initial velocity in the y-direction \( v_{iy} = 0 \) (because it's not mentioned), use:\[ v_y = v_{iy} + a_y t \]Substitute \( a_y = 7.00 \text{ m/s}^2 \) and the computed \( t \).
03

Find the magnitude of the velocity

The magnitude of the velocity \( v \) is given by the Pythagorean theorem:\[ v = \sqrt{v_x^2 + v_y^2} \]Use the values of \( v_x \) and \( v_y \) obtained in Step 2 to calculate \( v \).
04

Find the angle of the velocity

The angle \( \theta \) of the velocity vector with respect to the x-axis is given by:\[ \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) \]Use the values of \( v_x \) and \( v_y \) from Step 2 to find \( \theta \) in degrees.

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Key Concepts

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

Constant acceleration
In kinematics, constant acceleration occurs when the rate of change of velocity of an object remains the same throughout its motion. This means that with every passing second, the velocity of an object under constant acceleration changes by the same amount. The motion of objects with constant acceleration can be described with kinematic equations, which relate various aspects like velocity, displacement, and time.
For example, the acceleration vector \( \vec{a}=\left(5.00 \mathrm{\, m/s^{2}}\right) \hat{\mathrm{i}}+\left(7.00 \mathrm{\, m/s^{2}}\right) \hat{\mathrm{j}} \) describes constant rates of acceleration along both the x and y axes.
  • This indicates that the object is gaining speed at a steady rate of 5.00 m/s² to the right (x-direction) and 7.00 m/s² upwards (y-direction).
  • Understanding constant acceleration is pivotal because it simplifies the application of kinematic equations, allowing straightforward predictions about the object's future motion.
Overall, constant acceleration provides a reliable framework to predict the future state of a moving object using only a few known variables.
Velocity components
The concept of velocity components is crucial when dealing with objects moving in two or three dimensions. When an object moves at an angle or in a curved path, its velocity can be broken down into perpendicular components. Each component describes the velocity in a particular direction, typically along the x, y, and sometimes z axes in Cartesian coordinates.
For instance, given an initial velocity \( (4.00 \mathrm{\, m/s}) \hat{\mathbf{i}} \), the velocity is initially only in the x-direction. However, due to acceleration, the velocity also develops a y-component over time.
  • The x-component: This part of the velocity increases as a result of acceleration in the x-direction, using the formula \( v_x = v_{ix} + a_x t \).
  • The y-component: This initially absent component develops over time with acceleration in the y-direction, employing the formula \( v_y = v_{iy} + a_y t \).
By analyzing these components separately, we can totally understand the object's overall velocity at any given point in its trajectory.
Displacement
Displacement refers to the change in position of an object from its point of origin, and it is a vector quantity which means both magnitude and direction matter. In kinematics, displacement is often used to determine how far an object has traveled in a specific direction, irrespective of the path taken.
In the provided exercise, the pebble has a displacement of 12.0 m along the x-axis. This clear directional component of displacement helps us use kinematic equations easily to find the time of travel or other needed metric.
  • Utilizing the displacement equation \( s_x = v_{ix}t + \frac{1}{2} a_x t^2 \), we can relate displacement to time, initial velocity, and acceleration.
  • Displacement is frequently presented as the starting point for solving various aspects of motion, fundamentally framing the problem by pinning down a specific positional change.
Understanding displacement sets the stage for further calculations regarding velocity and time.
Quadratic equation
A quadratic equation is a polynomial equation of the second degree, generally expressed in the form \( ax^2 + bx + c = 0 \). Such equations often appear in kinematic problems when using equations of motion under constant acceleration.
In the context of the exercise, a quadratic equation arises when solving for time from the displacement equation \( 2.5t^2 + 4.00t - 12 = 0 \). To solve for time, the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) is used.
  • The quadratic formula allows for solving two possible values for time; however, only the positive value is physically meaningful in these scenarios since time cannot be negative.
  • Understanding and solving quadratic equations are fundamental skills not only in physics but also in various scientific and engineering disciplines.
Mastering these equations equips students with the tools to handle real-world problems involving motion and other forces.

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Most popular questions from this chapter

A projectile is fired horizontally from a gun that is \(45.0 \mathrm{~m}\) above flat ground, emerging from the gun with a speed of \(250 \mathrm{~m} / \mathrm{s}\). (a) How long does the projectile remain in the air? (b) At what horizontal distance from the firing point does it strike the ground? (c) What is the magnitude of the vertical component of its velocity as it strikes the ground?

Snow is falling vertically at a constant speed of \(8.0 \mathrm{~m} / \mathrm{s}\). At what angle from the vertical do the snowflakes appear to be falling as viewed by the driver of a car traveling on a straight, level road with a speed of \(50 \mathrm{~km} / \mathrm{h} ?\)

A person walks up a stalled 15 -m-long escalator in \(90 \mathrm{~s}\). When standing on the same escalator, now moving, the person is carried up in \(60 \mathrm{~s}\). How much time would it take that person to walk up the moving escalator? Does the answer depend on the length of the escalator?

At one instant a bicyclist is \(40.0 \mathrm{~m}\) due east of a park's flagpole, going due south with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\). Then \(30.0 \mathrm{~s}\) later, the cyclist is \(40.0 \mathrm{~m}\) due north of the flagpole, going due east with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\). For the cyclist in this \(30.0\) s interval, what are the (a) magnitude and (b) direction of the displacement, the (c) magnitude and (d) direction of the average velocity, and the (e) magnitude and (f) direction of the average acceleration?

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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