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Chapter 5: Problem 41

Particle Leaves Origin A particle leaves the origin with an initial velocity \(\vec{v}_{1}=(3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}\) and a constant acceleration \(\vec{a}=\) \(\left(-1.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(-0.500 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). When the particle reaches its maximum \(x\) coordinate, what are (a) its velocity and (b) its position vector?

Short Answer

Expert verified
(a) \( \vec{v} = -1.50 \mathrm{~m} / \mathrm{s} \hat{\mathrm{j}} \), (b) \( \vec{r} = 4.50 \mathrm{~m} \hat{\mathrm{i}} - 2.25 \mathrm{~m} \hat{\mathrm{j}} \).

Step by step solution

01

Understand the Problem

A particle starts at the origin with an initial velocity of \( \vec{v}_{1} = (3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}} \) and has a constant acceleration of \( \vec{a} = (-1.00 \mathrm{~m} / \mathrm{s}^{2}) \hat{\mathrm{i}} + (-0.500 \mathrm{~m} / \mathrm{s}^{2}) \hat{\mathrm{j}} \). The goal is to find the velocity and position vector when the particle reaches its maximum \( x \)-coordinate.
02

Determine Maximum x-Coordinate

To find the maximum x-coordinate, note that the velocity in the \( x \) direction will be zero at this point. The velocity function in the \( x \) direction is \( v_{x}(t) = v_{1x} + a_{x}t \). Set this to zero to solve for time \( t \). \[ 0 = 3.00 - 1.00t \Rightarrow t = 3.00 \mathrm{~s} \].
03

Find Velocity at Maximum x-Coordinate

Using the time calculated, determine the velocity in the \( y \) direction. The velocity function in the \( y \) direction is \( v_{y}(t) = v_{1y} + a_{y}t \). Given that the initial velocity in \( y \) direction is zero: \[ v_{y} = 0 + (-0.500)(3.00) = -1.50 \mathrm{~m} / \mathrm{s} \]. Hence, the velocity vector is \( \vec{v} = 0 \hat{\mathrm{i}} - 1.50 \hat{\mathrm{j}} \).
04

Find Position Vector

To find the position vector, use the position equations in the \( x \) and \( y \) directions. For the \( x \) direction: \[ x(t) = x_{0} + v_{1x}t + \frac{1}{2}a_{x}t^2 \Rightarrow x = 0 + (3.00)(3.00) + \frac{1}{2}(-1.00)(3.00)^2 \Rightarrow x = 4.50 \mathrm{~m} \]. For the \( y \) direction: \[ y(t) = y_{0} + v_{1y}t + \frac{1}{2}a_{y}t^2 \Rightarrow y = 0 + 0 + \frac{1}{2}(-0.500)(3.00)^2 \Rightarrow y = -2.25 \mathrm{~m} \]. The position vector is \( \vec{r} = 4.50 \hat{\mathrm{i}} - 2.25 \hat{\mathrm{j}} \).

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Key Concepts

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

Initial Velocity
Initial velocity describes the speed and direction at which a particle starts moving. In this problem, the particle leaves the origin with an initial velocity of \(\vec{v}_{1}=(3.00 \text{ m/s}) \hat{\mathbf{i}}\), which means it has a velocity of 3.00 meters per second in the positive x-direction. Understanding the initial velocity is crucial as it forms the starting point for analyzing the particle's subsequent motion under the influence of acceleration.
Constant Acceleration
Acceleration measures how the velocity of a particle changes over time. Here, the particle has a constant acceleration given by \(\vec{a} = (-1.00 \text{ m/s}^2) \hat{\mathbf{i}} + (-0.500 \text{ m/s}^2) \hat{\mathbf{j}}\). This tells us two things:
  • In the x-direction, the particle decelerates at 1.00 m/s虏.
  • In the y-direction, the particle accelerates at -0.500 m/s虏, indicating a downward direction.
This constant acceleration influences how the particle's velocity and position will change over time.
Position Vector
The position vector \(\vec{r}\) describes the location of a particle in space at any given time. It is found by combining the displacements in each direction.
For the x-direction, using the equation: \[ x(t) = x_{0} + v_{1x}t + \frac{1}{2}a_{x}t^2 \]
And for the y-direction: \[ y(t) = y_{0} + v_{1y}t + \frac{1}{2}a_{y}t^2 \]
At t = 3.00 s, the displacements are calculated as follows:
  • \(x = 4.50 \text{ m}\)
  • \(y = -2.25 \text{ m}\)
Hence, the position vector is \(\vec{r} = 4.50 \hat{\mathbf{i}} - 2.25 \hat{\mathbf{j}}\).
Velocity Vector
The velocity vector \(\vec{v}\) describes how fast and in which direction a particle is moving. It's calculated by combining the velocities in each direction.
For the x-direction: \[ v_{x}(t) = v_{1x} + a_{x}t \] and
For the y-direction: \[ v_{y}(t) = v_{1y} + a_{y}t \]
Given initial velocities and accelerations:
  • In the x-direction, at maximum x-coordinate, \(v_{x} = 0\)
  • In the y-direction, at t = 3.00 s, \(v_{y} = -1.50 \text{ m/s}\)
Thus, the velocity vector at maximum x-coordinate is \(\vec{v} = 0 \hat{\mathbf{i}} - 1.50 \hat{\mathbf{j}}\).
Kinematics Equations
Kinematics equations are essential for describing the motion of particles under constant acceleration. They relate displacement, velocity, acceleration, and time.
The key kinematics equations used in this problem are:
  • \(v = v_{0} + at\) (for velocity)
  • \(x = x_{0} + v_{0}t + \frac{1}{2}at^2\) (for position)
By applying these equations, we can find the maximum x-coordinate position and corresponding velocity and position vectors. They provide a systematic way to analyze and solve problems involving particle motion.

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Most popular questions from this chapter

A cat rides a merry-go-round while turning with uniform circular motion. At time \(t_{1}=2.00 \mathrm{~s}\), the cat's velocity is $$ \vec{v}_{1}=(3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}} $$ measured on a horizontal \(x y\) coordinate system. At time \(t_{2}=5.00 \mathrm{~s}\), its velocity is $$ \vec{v}_{2}=(-3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(-4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}} $$ What are (a) the magnitude of the cat's centripetal acceleration and (b) the cat's average acceleration during the time interval \(t_{2}-t_{1}\) ?

\(A\) rifle is aimed horizontally at a target \(30 \mathrm{~m}\) away. The bullet hits the target \(1.9 \mathrm{~cm}\) below the aiming point. What are (a) the bullet's time of flight and (b) its speed as it emerges from the rifle?

A particle moves so that its position as a function of time is \(\vec{r}(t)=(1 \mathrm{~m}) \hat{\mathrm{i}}+\left[\left(4 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}\right] \hat{\mathrm{j}}\). Write expressions for (a) its velocity and (b) its acceleration as functions of time.

Catapulted A stone is catapulted at Problem 6. time \(t_{1}=0\), with an initial velocity of magnitude \(20.0 \mathrm{~m} / \mathrm{s}\) and at an angle of \(40.0^{\circ}\) above the horizontal. What are the magnitudes of the (a) horizontal and (b) vertical components of its displacement from the catapult site at \(t_{2}=1.10\) s? Repeat for the (c) horizontal and (d) vertical components at \(t_{3}=1.80 \mathrm{~s}\), and for the (e) horizontal and (f) vertical components at \(t_{4}=5.00 \mathrm{~s}\).

A particle starts from the origin at \(t=0\) with a velocity of \(\vec{v}_{1}=(8.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) and moves in the \(x y\) plane with a constant acceleration of \(\vec{a}=\left(4.0 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(2.0 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). At the instant the particle's \(x\) coordinate is \(29 \mathrm{~m}\), what are (a) its \(y\) coordinate and (b) its speed?
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