/*! 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 17 A cart is propelled over an \(x ... [FREE SOLUTION] | 91Ó°ÊÓ

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A cart is propelled over an \(x y\) plane with acceleration components \(a_{x}=4.0 \mathrm{~m} / \mathrm{s}^{2}\) and \(a_{y}=-2.0 \mathrm{~m} / \mathrm{s}^{2}\). Its initial velocity has components \(v_{0 x}=8.0 \mathrm{~m} / \mathrm{s}\) and \(v_{0 y}=12 \mathrm{~m} / \mathrm{s}\). In unit-vector notation, what is the velocity of the cart when it reaches its greatest \(y\) coordinate?

Short Answer

Expert verified
The velocity at the greatest y-coordinate is \((32 \, \text{m/s}) \mathbf{i}\).

Step by step solution

01

Identify the Problem

We need to find the velocity of the cart in terms of unit-vector notation when it reaches its greatest y-coordinate. The given acceleration and initial velocity will be crucial in solving this.
02

Analyze the Y Motion

Since the greatest y-coordinate corresponds to the point where the upward velocity component becomes zero, set the final vertical velocity component to zero and solve for the time using the kinematic equation: \[ v_{y} = v_{0y} + a_{y}t \]Setting \( v_{y} = 0 \), we get:\[ 0 = 12 \, \text{m/s} + (-2 \, \text{m/s}^2)t \]
03

Solve for Time

Rearrange the equation from Step 2:\[ t = \frac{-v_{0y}}{a_{y}} = \frac{-12 \, \text{m/s}}{-2 \, \text{m/s}^2} = 6 \, \text{s} \]
04

Calculate the Final X-Component of Velocity

Now, use the time found in Step 3 to find the x-component of the velocity. Using the same type of kinematic equation:\[ v_{x} = v_{0x} + a_{x}t \]Substitute the values:\[ v_{x} = 8 \, \text{m/s} + (4 \, \text{m/s}^2)(6 \, \text{s}) = 32 \, \text{m/s} \]
05

Write the Final Velocity in Unit-Vector Notation

Since we know the final y-component of the velocity is 0 and the x-component as calculated is 32 m/s, the velocity in unit-vector notation is:\[ \mathbf{v} = (32 \, \text{m/s}) \mathbf{i} + (0 \, \text{m/s}) \mathbf{j} \]

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Key Concepts

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

Vector Components
Understanding vector components is fundamental in physics, especially when analyzing motion in two or three dimensions. Vectors have both magnitude and direction, and can be broken down into perpendicular components that align with the coordinate axes, typically the x and y axes.
  • The x-component represents the vector's horizontal influence.
  • The y-component represents the vertical influence.
By splitting vectors into components, complex motions can be simplified into a series of straightforward calculations. In this exercise, the initial velocity and acceleration are given in terms of their components, which allows for easier manipulation using kinematic equations.
Kinematic Equations
Kinematic equations are mathematical formulas used to describe the motion of objects under constant acceleration. These equations allow us to predict future motion, calculate the displacement, velocity, and time, and precisely understand how objects move.
The basic kinematic equations used in two-dimensional motion, where acceleration is constant, include:
  • \( v = v_0 + at \) - Final velocity after time \( t \).
  • \( s = s_0 + v_0t + \frac{1}{2}at^2 \) - Displacement after time \( t \).
  • \( v^2 = v_0^2 + 2as \) - Relates velocity and displacement.
These equations are particularly powerful when broken into components, as in this exercise, where we focus on the separate effects of acceleration on the x and y axes.
Acceleration
Acceleration is the rate at which an object's velocity changes. It is a vector quantity, meaning it has both magnitude and direction. Understanding how acceleration works is crucial in predicting the motion of any object.
In this exercise, the cart experiences two accelerations:
  • Positive acceleration in the x-direction: \( a_x = 4.0 \, \text{m/s}^2 \).
  • Negative acceleration in the y-direction: \( a_y = -2.0 \, \text{m/s}^2 \) (indicating downward motion).
By calculating the effects of these accelerations separately, we determine how the object's velocity changes over time in each direction. This helps to find the velocity of the cart when it reaches its highest point on the y-axis.
Velocity
Velocity refers to the rate at which an object changes its position. Like acceleration, velocity is a vector quantity and has both magnitude and direction. Calculating velocity involves understanding both initial velocity and changes due to acceleration.
In the given problem, the cart's initial velocity components are:
  • In the x-direction: \( v_{0x} = 8.0 \, \text{m/s} \).
  • In the y-direction: \( v_{0y} = 12.0 \, \text{m/s} \).
Using the kinematic equations, we calculate that the velocity in the y-direction becomes zero when the cart reaches the greatest y-coordinate, and the velocity in the x-direction increases to \( 32.0 \, \text{m/s} \) due to the positive acceleration on the x-axis.
Unit-Vector Notation
Unit-vector notation is a concise way to express vectors involving the standard unit vectors \( \mathbf{i} \) and \( \mathbf{j} \), which represent the x and y directions, respectively. This notation simplifies vector representation and calculations by breaking them into components along these axes.
For instance, in this exercise, the final velocity of the cart is expressed in unit-vector notation as:
  • \( \mathbf{v} = (32 \, \text{m/s}) \mathbf{i} + (0 \, \text{m/s}) \mathbf{j} \).
Here, \( \mathbf{i} \) is the horizontal direction, and there is no vertical component since the cart's y-velocity is zero at its highest point. Using unit-vector notation effectively communicates vector directions and magnitudes in physics, ensuring clarity and precision in problem-solving.

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Most popular questions from this chapter

The fast French train known as the TGV (Train à Grande Vitesse) has a scheduled average speed of \(216 \mathrm{~km} / \mathrm{h} .(\mathrm{a})\) If the train goes around a curve at that speed and the magnitude of the acceleration experienced by the passengers is to be limited to \(0.050 \mathrm{~g}\), what is the smallest radius of curvature for the track that can be tolerated? (b) At what speed must the train go around a curve with a \(1.00 \mathrm{~km}\) radius to be at the acceleration limit?

A woman who can row a boat at \(6.4 \mathrm{~km} / \mathrm{h}\) in still water faces a long, straight river with a width of \(6.4 \mathrm{~km}\) and a current of \(3.2 \mathrm{~km} / \mathrm{h}\). Let i point directly across the river and \(\hat{j}\) point directly downstream. If she rows in a straight line to a point directly opposite her starting position, (a) at what angle to i must she point the boat and (b) how long will she take? (c) How long will she take if, instead, she rows \(3.2 \mathrm{~km}\) down the river and then back to her starting point? (d) How long if she rows \(3.2 \mathrm{~km} u p\) the river and then back to her starting point? (e) At what angle to \(\hat{\mathrm{i}}\) should she point the boat if she wants to cross the river in the shortest possible time? (f) How long is that shortest time?

A boy whirls a stone in a horizontal circle of radius \(1.5 \mathrm{~m}\) and at height \(2.0 \mathrm{~m}\) above level ground. The string breaks, and the stone flies off horizontally and strikes the ground after traveling a horizontal distance of \(10 \mathrm{~m}\). What is the magnitude of the centripetal acceleration of the stone during the circular motion?

A ball is shot from the ground into the air. At a height of \(9.1 \mathrm{~m}\), its velocity is \(\vec{v}=(7.6 \hat{\mathrm{i}}+6.1 \mathrm{j}) \mathrm{m} / \mathrm{s}\), with \(\hat{\mathrm{i}}\) horizontal and upward. (a) To what maximum height does the ball rise? (b) What total horizontal distance does the ball travel? What are the (c) magnitude and (d) angle (below the horizontal) of the ball's velocity just before it hits the ground?

SSM ILW A particle leaves the origin with an initial velocity \(\vec{v}=(3,00 \hat{\mathrm{i}}) \mathrm{m} / \mathrm{s}\) and a constant acceleration \(\vec{a}=(-1.00 \hat{\mathrm{i}}-\) \(0 . .500 \hat{j}) \mathrm{m} / \mathrm{s}^{2}\), When it reaches its maximum \(x\) coordinate, what are its (a) velocity and (b) position vector?

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