/*! 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 motorbike starts from the orig... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 17

A motorbike starts from the origin and moves over an \(x y\) plane with acceleration components \(a_{x}=6.0 \mathrm{~m} / \mathrm{s}^{2}\) and \(a_{y}=-3.0 \mathrm{~m} / \mathrm{s}^{2}\). The initial velocity of the motorbike has components \(v_{0 x}=12.0 \mathrm{~m} / \mathrm{s}\) and \(v_{0 y}=18.0 \mathrm{~m} / \mathrm{s}\). Find the velocity of the motorbike, in unit-vector notation, when it reaches its greatest \(y\) coordinate.

Short Answer

Expert verified
The velocity is \(48.0 \hat{i} \, \text{m/s}\) when the motorbike reaches its greatest \(y\) coordinate.

Step by step solution

01

Understand the Problem

The motorbike is moving with constant acceleration components given by \(a_x = 6.0 \, \text{m/s}^2\) and \(a_y = -3.0 \, \text{m/s}^2\). We need to find its velocity in unit-vector notation when it reaches its highest point along the \(y\)-axis.
02

Identify When Greatest Y-Coordinate is Reached

The greatest \(y\) coordinate is reached when the velocity in the \(y\)-direction is zero \((v_y = 0)\). Since the initial velocity \(v_{0y} = 18.0 \, \text{m/s}\) and acceleration in \(y\) is \(a_y = -3.0 \, \text{m/s}^2\), we can calculate the time \(t\) it takes for \(v_y\) to become zero.
03

Calculate Time to Reach Greatest Y-Coordinate

Use the equation \( v_y = v_{0y} + a_y t \) and set \( v_y = 0 \) to solve for \( t \):\[ 0 = 18.0 + (-3.0) t \]\[ t = \frac{18.0}{3.0} = 6.0 \, \text{s} \]
04

Calculate Velocity in X-Direction at Time t

The velocity in the \(x\)-direction at time \(t\) is calculated using \( v_x = v_{0x} + a_x t \):\[ v_x = 12.0 + 6.0 \times 6.0 = 12.0 + 36.0 = 48.0 \, \text{m/s} \]
05

Write Final Velocity in Unit-Vector Notation

The velocity vector \(\mathbf{v}\) at the greatest \(y\) coordinate is:\[ \mathbf{v} = v_x \hat{i} + v_y \hat{j} = 48.0 \hat{i} + 0 \hat{j} \, \text{m/s} \]

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Key Concepts

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

Acceleration
Acceleration is a key concept in kinematics. It measures how quickly an object's velocity changes over time.
In our exercise, the motorbike experiences acceleration in two directions: horizontally and vertically.
It has horizontal acceleration component \(a_{x}=6.0 \, \text{m/s}^2\) and vertical acceleration component \(a_{y}=-3.0 \, \text{m/s}^2\).Here are some essential points about acceleration:
  • It can be constant or variable over time.
  • Positive acceleration means the object speeds up, while negative indicates slowing down.
  • In two dimensions, acceleration impacts both the "x" and "y" components of motion independently.
In this scenario, the motorbike speeds up in the "x" direction and slows down in the "y" direction. Acceleration is crucial in determining how long the motorbike takes to reach its peak "y" point, where it would temporarily halt vertical movement.
Velocity
Velocity is the rate at which an object changes its position. It is a vector quantity, having both magnitude and direction.
Unlike speed, velocity describes how fast and in what direction an object is moving.
Initially, the motorbike's velocity has components of \(v_{0x}=12.0 \, \text{m/s}\) and \(v_{0y}=18.0 \, \text{m/s}\).Some important aspects of velocity include:
  • It is possible to have zero velocity in one direction while maintaining non-zero in another, as seen in our problem.
  • When an object reaches its maximum height during upward motion, the vertical velocity (\(v_y\)) becomes zero.
  • Just like acceleration, velocity also has separate "x" and "y" components in two-dimensional motion.
In the task, when the bike reaches its highest vertical point, it still has a horizontal velocity of \(48 \, \text{m/s}\), indicating ongoing movement along the "x" axis.
Unit-Vector Notation
Unit-vector notation is a method for expressing vectors in terms of their components along the standard unit vectors, typically \(\hat{i}\) for the "x" direction and \(\hat{j}\) for the "y" direction.
This notation helps in clearly specifying the exact horizontal and vertical influences on motion.
Some core aspects of unit-vector notation include:
  • It breaks down any vector quantity into its constituent components which are easily handled mathematically.
  • In our example, the final velocity is expressed as \(\mathbf{v} = 48.0 \hat{i} + 0 \hat{j} \, \text{m/s}\).
  • It simplifies calculating resultant vectors in physics by aligning components along orthogonal axes.
Using unit-vector notation ensures a clear and precise representation of velocities, especially in a multi-directional motion like in this exercise.
Two-Dimensional Motion
The concept of two-dimensional motion addresses movement that occurs in both "x" and "y" directions simultaneously.
This type of motion is often seen in real-world scenarios, such as projectile motion or a vehicle traveling across a plane.
In this exercise, the motorbike moves in a 2D plane with changing velocities and accelerations in each direction. Key points about two-dimensional motion involve:
  • It requires analyzing motion in two perpendicular directions independently.
  • Kinematic equations can be applied separately to each direction to find velocities, distances, or times.
  • Results from both directions are often combined to understand the overall trajectory or movement path of the object completely.
In solving exercises involving 2D motion, understanding the unique behavior of each component helps predict and calculate the object's overall dynamics accurately.

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Most popular questions from this chapter

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?

Snow is falling vertically at a constant speed of \(8.0 \mathrm{~m} / \mathrm{s}\). At what angle from the vertical do the snowflakes appear to be falling as viewed by the driver of a car traveling on a straight, level road with a speed of \(50 \mathrm{~km} / \mathrm{h}\) ?

When a large star becomes a supernova, its core may be compressed so tightly that it becomes a neutron star, with a radius of about \(20 \mathrm{~km}\) (about the size of the San Francisco area). If a neutron star rotates once every second, (a) what is the speed of a particle on the star's equator and (b) what is the magnitude of the particle's centripetal acceleration? (c) If the neutron star rotates faster, do the answers to (a) and (b) increase, decrease, or remain the same?

The acceleration of a particle moving only on a horizontal \(x y\) plane is given by \(\vec{a}=3 t \hat{\mathrm{i}}+4 t \hat{\mathrm{j}}\), where \(\vec{a}\) is in meters per secondsquared and \(t\) is in seconds. At \(t=0\), the position vector \(\vec{r}=(20.0 \mathrm{~m}) \hat{\mathrm{i}}+(40.0 \mathrm{~m}) \hat{\mathrm{j}}\) locates the particle, which then has the velocity vector \(\vec{v}=(5.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(2.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). At \(t=4.00 \mathrm{~s}\), what are (a) its position vector in unit-vector notation and (b) the angle between its direction of travel and the positive direction of the \(x\) axis?

Two seconds after being projected from ground level, a projectile is displaced \(40 \mathrm{~m}\) horizontally and \(58 \mathrm{~m}\) vertically above its launch point. What are the (a) horizontal and (b) vertical components of the initial velocity of the projectile? (c) At the instant the projectile achieves its maximum height above ground level, how far is it displaced horizontally from the launch point?
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