Chapter 4: Problem 15
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
Step by step solution
Understand the motion in the x-direction
Calculate the time to reach maximum x-coordinate
Find the velocity in the y-direction at that time
Determine the velocity vector at maximum x-coordinate
Calculate the x-coordinate at maximum x
Calculate the y-coordinate at maximum x
Determine the position vector at maximum x-coordinate
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Projectile Motion
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.
Acceleration
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.
Velocity
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 \(v_y = v_{0y} + a_y \, t\).
Position Vectors
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.