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From the origin, a particle starts at \(t=0 \mathrm{~s}\) with a velocity \(\vec{v}=7.0 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}\) and moves in the \(x y\) plane with a constant acceleration of \(\vec{a}=(-9.0 \hat{\mathrm{i}}+3.0 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}^{2} .\) At the time the particle reaches the maximum \(x\) coordinate, what is its (a) velocity and (b) position vector?

Short Answer

Expert verified
Velocity: \(0 \hat{i} + 2.333 \hat{j} \text{ m/s}\); Position: \(2.722 \hat{i} + 0.815 \hat{j} \text{ m}\).

Step by step solution

01

Understand the motion in the x-direction

We know the initial velocity in the x-direction is \(v_{0x} = 7.0 \ \text{m/s}\) and the acceleration in the x-direction is \(a_x = -9.0 \ \text{m/s}^2\). The particle reaches the maximum x-coordinate when its x-velocity is zero.
02

Calculate the time to reach maximum x-coordinate

Set the velocity in the x-direction to zero to find the time: \[v_x = v_{0x} + a_x t = 0 \Rightarrow 7.0 + (-9.0)t = 0 \Rightarrow t = \frac{7.0}{9.0} \, \text{s} \approx 0.778 \ \text{s}\]!.
03

Find the velocity in the y-direction at that time

The velocity in the y-direction can be calculated using \(v_y = v_{0y} + a_y t\). Here, \(v_{0y} = 0 \text{ m/s}\) and \(a_y = 3.0 \text{ m/s}^2\). \[v_y = 0 + 3.0 \times \frac{7.0}{9.0} = 2.333 \text{ m/s}\].
04

Determine the velocity vector at maximum x-coordinate

The velocity vector when \(v_x = 0\) and \(v_y\) is calculated as above, is:\[\vec{v} = 0 \hat{i} + 2.333 \hat{j} \text{ m/s}\].
05

Calculate the x-coordinate at maximum x

Use the equation \(x = v_{0x} t + \frac{1}{2} a_x t^2\) to find the maximum x-coordinate:\[x = 7.0 \cdot \frac{7.0}{9.0} + \frac{1}{2} (-9.0) \left(\frac{7.0}{9.0}\right)^2\] \[x = 2.722 \text{ m}\].
06

Calculate the y-coordinate at maximum x

Use the equation \(y = v_{0y} t + \frac{1}{2} a_y t^2\) to find the y-coordinate:\[y = 0 + \frac{1}{2} (3.0) \left(\frac{7.0}{9.0}\right)^2\] \[y = 0.815 \text{ m}\].
07

Determine the position vector at maximum x-coordinate

The position vector at the maximum x-coordinate is:\[\vec{r} = 2.722 \hat{i} + 0.815 \hat{j} \text{ m}\].

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Key Concepts

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

Projectile Motion
Projectile motion refers to the two-dimensional movement of an object under the influence of gravity alone. In the absence of air resistance, an object in projectile motion will follow a parabolic path. This path is determined by its initial velocity and the constant acceleration due to gravity acting on the object.
In the given exercise, the particle moves in the xy-plane with an initial velocity and a constant acceleration. Although the motion in the x and y directions is interconnected, they can be analyzed independently.
Key points about projectile motion include:
  • Two Dimensional Movement: Projectile motion takes place in a plane, often referred to as the xy-plane. This requires breaking down motion into horizontal (x-axis) and vertical (y-axis) components.
  • Independence of Axes: The horizontal and vertical components of projectile motion operate independently of each other. Horizontal movements do not influence vertical movements and vice versa.
  • Parabolic Trajectory: The object's path is parabolic due to the influence of gravity acting on it vertically.
Understanding these principles helps you break down complex problems like the one described, allowing you to solve for different variables such as time, velocity, or position at particular points along the path.
Acceleration
Acceleration is the rate of change in velocity of an object over time. In the context of projectile motion, it plays a vital role in influencing the motion's trajectory. For the exercise at hand, we have a constant acceleration vector given as \(-9.0 \hat{i} + 3.0 \hat{j}\ m/s^2\). This vector tells us a lot about how the object's velocity changes over time.
Here's how to break it down:
  • Acceleration in the x-direction: \(a_x = -9.0 m/s^2\) implies that the velocity in the x-direction is decreasing over time, as the negative sign indicates a deceleration.
  • Acceleration in the y-direction: \(a_y = 3.0 m/s^2\) indicates that the y-velocity is increasing, hence the object gains speed in the vertical direction over time.
The acceleration affects how fast the particle moves and the ultimate shape of its trajectory. Constant acceleration like in this problem simplifies calculations, as it allows for the use of kinematic equations.
Velocity
Velocity is a vector quantity that includes both the speed and direction of an object's movement. The velocity of an object in projectile motion can change due to acceleration.
In our exercise, the initial velocity is \(7.0 \hat{i} m/s\). Let's see how it plays out in the x and y directions:
  • X-Direction Velocity: The initial velocity in the x-direction is given by \(v_{0x} = 7.0 m/s\), and it decreases over time because of the negative x-acceleration. When it becomes zero, the particle reaches its maximum x-coordinate.
  • Y-Direction Velocity: Initially, the y-direction velocity is zero \(v_{0y} = 0 m/s\), but it increases over time under positive y-acceleration. This is calculated using the formula \(v_y = v_{0y} + a_y \, t\).
At the point where the x-direction velocity is zero, the object's velocity vector is purely vertical. Understanding how velocity in each direction changes with time helps in finding other characteristics of the particle's path, such as its position at any given instant.
Position Vectors
The position vector of a particle provides a way to represent its exact location in the xy-plane at any given time. In projectile motion, this vector combines both x and y coordinates. It is often presented in the form \( \vec{r} = x \hat{i} + y \hat{j}\).
In this exercise, the process of finding the position vector involves:
  • X-Coordinate Calculation: The x-coordinate is determined using the equation: \(x = v_{0x} t + \frac{1}{2} a_x t^2\). This gives the actual horizontal position at any time t.
  • Y-Coordinate Calculation: Similarly, the y-coordinate can be found using: \(y = v_{0y} t + \frac{1}{2} a_y t^2\). This provides the vertical position.
At the maximum x-coordinate in the exercise, these calculations give the position vector \(\vec{r} = 2.722 \hat{i} + 0.815 \hat{j} \, m\). By assessing both the calculated x and y components, we understand the precise location of the particle in its path.

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