/*! 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 2 \Lambda particle moving in \(x-y... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\Lambda particle moving in \(x-y\) plane is at origin at time \(t=0 .\) Velocity of the particle is \((2 \vec{i}-4 \hat{j}) \mathrm{m} / \mathrm{s}\) and acceleration \(i s(4 \hat{i} \mid \hat{j}) \mathrm{m} / \mathrm{s}^{2}\). Find at \(t=2 \mathrm{~s}\) (a) velocity of particle and (b) coordinates of particle.

Short Answer

Expert verified
(a) Velocity: \(10\hat{i} - 2\hat{j}\) m/s, (b) Coordinates: \((12, -6)\).

Step by step solution

01

Understand the Problem

The problem gives us the initial velocity and acceleration of a Lambda particle moving in the x-y plane. We're asked to find the velocity and coordinates of the particle after 2 seconds.
02

Determine Velocity at t = 2s

Use the formula for velocity as a function of time: \( \vec{v}(t) = \vec{v}_0 + \vec{a}t \). Here, \( \vec{v}_0 = 2\hat{i} - 4\hat{j} \) m/s and \( \vec{a} = 4\hat{i} + \hat{j} \) m/s². Substitute these values and \( t = 2 \) s: \[ \vec{v}(2) = (2\hat{i} - 4\hat{j}) + (4\hat{i} + \hat{j})(2) \] \[ \vec{v}(2) = 2\hat{i} - 4\hat{j} + 8\hat{i} + 2\hat{j} \]\[ \vec{v}(2) = 10\hat{i} - 2\hat{j} \] m/s.
03

Calculate Position at t = 2s

Use the formula for position as a function of time: \( \vec{r}(t) = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2}\vec{a}t^2 \). Here, \( \vec{r}_0 = 0 \) (origin), \( \vec{v}_0 = 2\hat{i} - 4\hat{j} \), \( \vec{a} = 4\hat{i} + \hat{j} \), and \( t = 2 \) s. Calculate:\[ \vec{r}(2) = 0 + (2\hat{i} - 4\hat{j})(2) + \frac{1}{2}(4\hat{i} + \hat{j})(4) \]\[ \vec{r}(2) = (4\hat{i} - 8\hat{j}) + (8\hat{i} + 2\hat{j}) \]\[ \vec{r}(2) = 12\hat{i} - 6\hat{j} \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Kinematics
Kinematics is a branch of physics that deals with the motion of objects without considering the forces that cause the motion. It focuses on describing the motion through the relationship between displacement, velocity, and acceleration.

The key elements of kinematics include:
  • Displacement: A vector quantity that represents the change in position of an object.
  • Velocity: The rate of change of an object's displacement over time. It's a vector quantity, meaning it has both magnitude and direction.
  • Acceleration: The rate of change of velocity over time. It also is a vector quantity.
Understanding kinematics allows you to predict an object's future position and velocity using mathematical equations, even if you don't know why the object is moving that way.
Velocity Calculation
Velocity is a crucial concept in the study of motion. It describes how fast an object is moving and in what direction. The formula to calculate the velocity of an object, especially when considering time-dependent acceleration, is:

  • Initial velocity (\(\vec{v}_0\):
  • Acceleration (\(\vec{a}\)) and time (\(t\)) terms:
  • Final velocity (\(\vec{v}(t)\) can be found using:
    \[ \vec{v}(t) = \vec{v}_0 + \vec{a}t \]
In this exercise, the Lambda particle's initial velocity was given by \(2\hat{i} - 4\hat{j}\) m/s. The acceleration was \(4\hat{i} + \hat{j}\) m/s². Thus, to find its velocity at \(t = 2\) seconds, you simply substitute these values into the formula, getting:\[ \vec{v}(2) = (2\hat{i} - 4\hat{j}) + (4\hat{i} + \hat{j})(2) = 10\hat{i} - 2\hat{j} \] m/s.

This result provides both the speed and the direction of the particle at this time.
Position Vector
The position vector describes the location of an object as it moves in the coordinate plane. This vector is crucial for determining where an object is located at any given time. It considers initial position, velocity, and acceleration.

The equation to determine the position vector is:
  • Initial position (\(\vec{r}_0\)) -- In this case, it starts at the origin:
  • Velocity (\(\vec{v}_0\)) and time (\(t\)) terms:
  • Acceleration (\(\vec{a}\) can be factored into the formula:
    \[ \vec{r}(t) = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2}\vec{a}t^2 \]
For this problem, substituting the known values, you get:\[ \vec{r}(2) = 0 + (2\hat{i} - 4\hat{j})(2) + \frac{1}{2}(4\hat{i} + \hat{j})(4) = 12\hat{i} - 6\hat{j} \]This calculation helps determine the coordinates of the Lambda particle at \(t = 2\) seconds, showing exactly where the particle will be on the \(x-y\) plane.
Acceleration
Acceleration is an essential component when analyzing motion, as it captures how the velocity of an object changes over time. Unlike velocity, which describes speed and direction, acceleration tells you how quickly the velocity itself is changing.

Acceleration is calculated based on:
  • Initial velocity (\(\vec{v}_0\))
  • Final velocity (\(\vec{v}(t)\)) -- determined over a specific time duration
  • Time interval (\(t\))
The acceleration for the Lambda particle is constant and given by \(4\hat{i} + \hat{j}\) m/s². This constant acceleration allows us to express both velocity and position as linear functions of time. The continuous change imposed by acceleration is what causes the particle to shift its velocity and ultimately its position on the \(x-y\) plane, allowing us to predict future motion diligently with mathematical precision.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The maximum height attained by the projectile depends upon the following factors: (a) magnitude of initial velocity (b) acceleration of the projectile (c) angle of projection (d) mass of the projectile

A particle is projecled from a horizontal plane with speed \(u\) at some ungle. At highest point its velocity is found to be \(u / 2\). The maximum height of the projcctile will be (a) \(\frac{u^{2}}{4 g}\) (b) \(\frac{3 u^{2}}{4 g}\) (c) \(\frac{3 u^{2}}{8 g}\) (d) \(\frac{u^{2}}{8 g}\)

Two bullets are thrown simultaneously in the same vertical plane from the same point with different speeds \(v_{1}\) and \(v_{2}\) making angles \(\theta_{1}\) and \(\theta_{2}\) respectively with the horizontal such that \(v_{1} \cos \theta_{1}=v_{2} \cos \theta_{2}\) The path followed by one as seen by the other is: (a) a vertical straight line (b) a horizontal straight line (c) a parabola (d) None

\Lambdat time \(t=0\), a particle is at \((2 \mathrm{~m}, 4 \mathrm{~m})\). It starts moving towards positive \(x\) -axis with constant acceleration \(2 \mathrm{~m} / \mathrm{s}^{2}\) (initial velocity \(=0\) ). \(\Lambda\) fter \(2 \mathrm{~s}\) an acceleration of \(4 \mathrm{~m} / \mathrm{s}^{2}\) starts acting on the particle in negative \(y\) -direction also. Find after next \(2 \mathrm{~s}\) : (a) velocity and (b) coordinates of particle.

Statement-1 : In projectile motion if angle of projection is made \(60^{\circ}\) instead of \(30^{\circ}\), the minimum kinctie cnergy will become \(\frac{1}{3}\) limes. Initial speed remaining the same. Statement-2 : In projectile motion, minimum kinetic energy is proportional to \(\cos ^{2} \theta\), where \(\theta\) is angle of projection.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.