/*! 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 5 A particle passes through the or... [FREE SOLUTION] | 91Ó°ÊÓ

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A particle passes through the origin with a velocity of \((6.2 \mathrm{m} / \mathrm{s}) \hat{\mathrm{y}}\). If the particle's acceleration is \(\left(-4.4 \mathrm{m} / \mathrm{s}^{2}\right) \hat{\mathrm{x}}\) (a) what are its \(x\) and \(y\) positions after 5.0 s? (b) What are \(v_{x}\) and \(v_{y}\) at this time? \((c)\) Does the speed of this particle increase with time, decrease with time, or increase and then decrease? Explain.

Short Answer

Expert verified
(a) x = -55.0 m, y = 31.0 m (b) v_x = -22.0 m/s, v_y = 6.2 m/s (c) The speed increases with time.

Step by step solution

01

Determine the y-position

The particle's motion in the y-direction is characterized by a constant velocity, as there is no acceleration. The initial position in the y-direction is zero and the velocity is given as 6.2 m/s. The position in the y-direction at time \( t \) is given by the equation: \( y = v_{y} imes t = 6.2 \, \mathrm{m/s} \times 5.0 \, \mathrm{s} = 31.0 \, \mathrm{m} \). Thus, the y-position after 5.0 seconds is 31.0 m.
02

Determine the x-position

The particle's motion in the x-direction involves acceleration. We start from rest, so the velocity is zero except for acceleration. The initial velocity \( v_{x0} \) = 0 and acceleration \( a_{x} = -4.4 \, \mathrm{m/s^2} \). The x-position is given by the equation: \( x = v_{x0} \, t + \frac{1}{2} a_{x} \, t^2 = 0 + \frac{1}{2} \times (-4.4) \times (5.0)^2 = -55.0 \, \mathrm{m} \).Hence, the x-position after 5.0 seconds is -55.0 m.
03

Determine velocities in x and y directions

For the y-direction, the velocity remains constant as there is no acceleration, so \( v_y = 6.2 \, \mathrm{m/s} \).In the x-direction, the velocity changes due to constant acceleration from rest. Using the equation \( v_x = v_{x0} + a_x \, t \), we have:\( v_x = 0 + (-4.4) \, \mathrm{m/s^2} \times 5.0 \, \mathrm{s} = -22.0 \, \mathrm{m/s} \).So the velocities at 5.0 seconds are \( v_x = -22.0 \, \mathrm{m/s} \) and \( v_y = 6.2 \, \mathrm{m/s} \).
04

Determine if speed increases or decreases

Speed is the magnitude of the velocity vector given by \( |v| = \sqrt{v_x^2 + v_y^2} \). Initially, \( v_x = 0 \) and \( v_y = 6.2 \, \mathrm{m/s} \) so the initial speed is 6.2 m/s. After 5.0 seconds, \( v_x = -22.0 \, \mathrm{m/s} \) and \( v_y = 6.2 \, \mathrm{m/s} \), so: \( |v| = \sqrt{(-22.0)^2 + (6.2)^2} = \sqrt{484 + 38.44} = \sqrt{522.44} \approx 22.85 \, \mathrm{m/s} \).Thus, the speed increases with time, even though the acceleration affects only the x-direction.

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Key Concepts

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

Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause this motion.
In kinematics, we typically analyze the motion through parameters such as displacement, velocity, and acceleration.
This allows us to create a clear picture of how objects move through space over time.

We primarily use kinematic equations to determine various aspects of a particle's movement based on initial conditions like velocity and acceleration.
For example, when we're interested in the positions along the x and y axes, kinematics helps us understand how these positions change over specified time intervals.
  • In the y-direction, a constant velocity implies a straightforward calculation for position.
  • Conversely, in the x-direction, acceleration plays a crucial role in determining the particle's future position.
This leads us to solve problems involving motion in two dimensions, where each axis might be subject to different conditions.
Velocity
Velocity is a vector quantity, meaning it has both magnitude and direction.
Understanding velocity is key to analyzing motion, especially in two dimensions where each direction can have its own velocity component.
  • The y-velocity component in our exercise remains constant at 6.2 m/s, as no acceleration affects it.
  • On the other hand, the x-velocity changes because it is influenced by an acceleration of -4.4 m/s².
To find the component of velocity at any point in time, we use the equation:
\[ v_x = v_{x0} + a_x \, t \]
In this setting, the x-velocity changes considerably over 5 seconds, turning from 0 at the start to -22 m/s, indicating a reverse in direction.
This demonstrates how acceleration influences velocity when present.
Acceleration
Acceleration is the rate at which an object's velocity changes with time.
It is also a vector quantity, able to affect various directional components of a particle's journey.

In the exercise at hand, the particle experiences acceleration only in the x-direction with a value of -4.4 m/s².
This means each second, the x-component of velocity changes by -4.4 m/s.
  • Since the y-direction has no acceleration, its velocity remains constant throughout.
  • The main effect of acceleration in the x-direction is to reduce the velocity as time progresses, showing a dynamic change in motion.
The concept of acceleration is crucial because it not only changes velocities but also affects how and where an object moves over time.
Position-Time Relationship
The position-time relationship in kinematics describes how an object's position changes over time based on its velocity and acceleration.
In our specific situation, understanding this relationship involves analyzing both the x and y directions separately.

As time progresses, each position component evolves according to its initial conditions and influences:
  • For the y-direction: \[ y = v_{y} \times t \] The absence of acceleration implies that position changes linearly with time.
  • For the x-direction: \[ x = v_{x0} \, t + \frac{1}{2} a_{x} \, t^2 \] Here, acceleration modifies the position significantly over time.
This relationship is the core of analyzing motion, as it lets us predict where the particle will be at any given moment.

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Most popular questions from this chapter

IP Snowballs are thrown with a speed of \(13 \mathrm{m} / \mathrm{s}\) from a roof \(7.0 \mathrm{m}\) above the ground. Snowball \(\mathrm{A}\) is thrown straight downward; snowball \(\mathrm{B}\) is thrown in a direction \(25^{\circ}\) above the horizontal. (a) Is the landing speed of snowball A greater than, less than, or the same as the landing speed of snowball B? Explain. (b) Verify your answer to part (a) by calculating the landing speed of both snowballs.

A ball rolls off a table and falls \(0.75 \mathrm{m}\) to the floor, landing with a speed of \(4.0 \mathrm{m} / \mathrm{s}\). (a) What is the acceleration of the ball just before it strikes the ground? (b) What was the initial speed of the ball? (c) What initial speed must the ball have if it is to land with a speed of \(5.0 \mathrm{m} / \mathrm{s} ?\)

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Suppose we change the dolphin's launch angle to \(45.0^{\circ}\), but everything else remains the same. Thus, the horizontal distance to the ball is \(5.50 \mathrm{m}\), the drop height is \(4.10 \mathrm{m},\) and the dolphin's launch speed is \(12.0 \mathrm{m} / \mathrm{s}\). (a) What is the vertical distance between the dolphin and the ball when the dolphin reaches the horizontal position of the ball? We refer to this as the "miss distance." (b) If the dolphin's launch speed is reduced, will the miss distance increase, decrease, or stay the same? (c) Find the miss distance for a launch speed of \(10.0 \mathrm{m} / \mathrm{s}\).

A particle leaves the origin with an initial velocity \(\overrightarrow{\mathbf{v}}=(2.40 \mathrm{m} / \mathrm{s}) \hat{\mathrm{x}},\) and moves with constant acceleration \(\overrightarrow{\mathbf{a}}=\left(-1.90 \mathrm{m} / \mathrm{s}^{2}\right) \hat{\mathrm{x}}+\left(3.20 \mathrm{m} / \mathrm{s}^{2}\right) \hat{\mathrm{y}}\) (a) How far does the particle move in the \(x\) direction before turning around? (b) What is the particle's velocity at this time? (c) Plot the particle's position at \(t=0.500 \mathrm{s}, 1.00 \mathrm{s}, 1.50 \mathrm{s},\) and \(2.00 \mathrm{s}\). Use these results to sketch position versus time for the particle.

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