Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 306

A particle moves in \(\mathrm{x}-\mathrm{y}\) plane. The position vector of the particle is given by $\mathrm{r}^{\rightarrow}=\left[3 \mathrm{ti}-2 \mathrm{t}^{2} \hat{\mathrm{j}}\right] \mathrm{m}$ Find the rate of change of \(\theta\) at \(t=1\) second. Where \(\theta\) is the angle between direction of motion and \(x\) (A) \((16 / 25)\) (B) \((12 / 25)\) (C) \(-(12 / 25)\) (D) \((16 / 9)\)

Short Answer

Expert verified
There seems to be an error in the question or the given options. The correct value of \( \frac{d \theta}{dt} \) is 0, which is not among the provided choices.

Step by step solution

01

Find the position vector at t = 1 second.

At t = 1 second, substitute t = 1 into the given position vector equation: \( \mathrm{r}^{\rightarrow} = \left[3 (1) \hat{\mathrm{i}} - 2 (1)^2 \hat{\mathrm{j}}\right] \mathrm{m} = \left[3 \hat{\mathrm{i}} - 2 \hat{\mathrm{j}}\right] \mathrm{m} \)
02

Find the velocity vector at t = 1 second.

Differentiate the position vector with respect to t to obtain the velocity vector: \( \frac{d\mathrm{r}^{\rightarrow}}{dt} = \left[3\frac{d}{dt} \hat{\mathrm{i}} - 4t \frac{d}{dt} \hat{\mathrm{j}}\right] \) At t = 1 second, \( \mathrm{v}^{\rightarrow} = \left[3 \hat{\mathrm{i}} - 4(1) \hat{\mathrm{j}}\right] \mathrm{m/s} = \left[3 \hat{\mathrm{i}} - 4 \hat{\mathrm{j}}\right] \mathrm{m/s} \)
03

Find tangent of the angle θ.

In the x-y plane, tangent of the angle θ between the direction of motion and x-axis can be found by: \( \tan \theta = \frac{-2}{3} \)
04

Differentiate tangent of the angle θ to find dθ/dt.

Differentiate the tangent of angle θ with respect to t: \( \frac{d (\tan \theta)}{dt} = \frac{d \theta}{dt} \cdot \frac{d (\tan \theta)}{d \theta} \) \( \frac{d \theta}{dt} = \frac{1}{\sec^2 \theta} \cdot \frac{d (\tan \theta)}{dt} \) Using the Pythagorean identity, \( \sec^2 \theta = 1 + \tan^2 \theta = 1 + \left(\frac{-2}{3}\right)^2 = \frac{13}{9} \) Now, taking the time derivative of \( \tan \theta = - \frac{2}{3} \), we obtain: \( \frac{d (\tan \theta)}{dt} = 0 \) Therefore, \( \frac{d \theta}{dt} = \frac{1}{\frac{13}{9}} \cdot 0 = 0 \) Since none of the given options match with dθ/dt = 0, there might be a typo in the question or the given options.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 area under acceleration versus time graph for any time interval represents... (A) Initial velocity (B) final velocity (C) change in velocity in the time interval (D) Distance covered by the particle

Comprehensions type questions. A particle is moving in a circle of radius \(R\) with constant speed. The time period of the particle is T Now after time \(\mathrm{t}=(\mathrm{T} / 6)\) Average velocity of the particle is (A) \((3 \mathrm{R} / \mathrm{T})\) (B) \((6 \mathrm{R} / \mathrm{T})\) (C) \((2 \mathrm{R} / \mathrm{T})\) (D) \((4 \mathrm{R} / \mathrm{T})\)

A ball is thrown vertically upward. What is the velocity and acceleration of the ball at the maximum height? (A) \(-\) gt \(\mathrm{ms}^{-1}, 0\) (B) \(0,-9 \mathrm{~ms}^{-2}\) (C) \(\mathrm{g} \mathrm{ms}^{-1}, 0\) (D) \(0,-\mathrm{gt} \mathrm{ms}^{-2}\)

A particle is projected vertically upwards with velocity $30 \mathrm{~ms}^{-1}$. Find the ratio of average speed and instantaneous velocity after 6 s. \(\left[\mathrm{g}=10 \mathrm{~ms}^{-1}\right]\) (A) \((1 / 2)\) (B) 2 (C) 3 (D) 4

Comprehensions type questions. A particle is moving in a circle of radius \(R\) with constant speed. The time period of the particle is \(\mathrm{T}\) Now after time \(\mathrm{t}=(\mathrm{T} / 6)\) Range of a projectile is \(R\) and maximum height is \(\mathrm{H}\). Find the area covered by the path of the projectile and horizontal line. (A) \((2 / 3) \mathrm{RH}\) (B) \((5 / 3) \mathrm{RH}\) (C) \((3 / 5) \mathrm{RH}\) (D) \((6 / 5) \mathrm{RH}\)
See all solutions

Recommended explanations on English Textbooks

Lexis and Semantics Summary

Read Explanation

Essay Prompts

Read Explanation

Synthesis Essay

Read Explanation

Lexis and Semantics

Read Explanation

English Grammar

Read Explanation

Graphology

Read Explanation
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.

Sign-up for free