Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 213

An object moves in a straight line. It starts from the rest and its acceleration is \(2 \mathrm{~ms}^{-2}\). After reaching a certain point it comes back to the original point. In this movement its acceleration is $-3 \mathrm{~ms}^{-2}$. till it comes to rest. The total time taken for the movement is 5 second. Calculate the maximum velocity. (A) \(6 \mathrm{~ms}^{-1}\) (B) \(5 \mathrm{~ms}^{-1}\) (C) \(10 \mathrm{~ms}^{-1}\) (D) \(4 \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
The maximum velocity is \(6 \mathrm{~ms}^{-1}\) (A).

Step by step solution

01

Write equations for displacement for each part of motion

Since the object comes back to the original position, the displacement in each part of the motion must be equal and opposite. Using the equation: \(x = ut + \frac{1}{2}at^2\), we have: For the first part of the motion: \(x = 0 + \frac{1}{2}(2)t_1^2\) For the second part of the motion: \(-x = 0 + \frac{1}{2}(-3)t_2^2\)
02

Solve for the time taken in each part of the motion

Using the given information t1 + t2 = 5 s, and equating the magnitudes of displacement from the previous step, we have: \(t_1^2 = \frac{3}{2}t_2^2\) Substituting t2 = 5 - t1, we get: \(t_1^2 = \frac{3}{2}(5 - t_1)^2\) Solve for t1: \(t_1 \approx 3 \mathrm{s}\) So, t2 = 5 - t1 = 2 s.
03

Find maximum velocity

Now we know the time taken for the first part of the motion, we can calculate the maximum velocity using the initial velocity and acceleration for that part: \(v_\text{max} = u_1 + a_1t_1\) Since the object starts from rest, u1 = 0. The acceleration for the first part of the motion, a1 = 2 ms^-2, and the time taken, t1 = 3 s. Thus, we can plug in these values to find the maximum velocity: \(v_\text{max} = 0 + (2)(3)\) \(v_\text{max} = 6 \mathrm{~ms}^{-1}\) Therefore, the maximum velocity is \(6 \mathrm{~ms}^{-1}\) and the correct answer is (A) \(6 \mathrm{~ms}^{-1}\).

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

A particle has initial velocity \((2 \hat{1}+3 \hat{j}) \mathrm{ms}^{-1}\) and has acceleration \((\hat{1}+\hat{j}) \mathrm{ms}^{-2}\). Find the velocity of the particle after 2 second. (A) \((3 \hat{1}+5 \hat{j}) \mathrm{ms}^{-1}\) (B) \((4 \hat{i}+5 \hat{\jmath}) \mathrm{ms}^{-1}\) (C) \((3 \hat{1}+2 \hat{j}) \mathrm{ms}^{-1}\) (D) \((5 \hat{1}+4 \hat{j}) \mathrm{ms}^{-1}\)

A particle moves in straight line. Its position is given by $\mathrm{x}=2+5 \mathrm{t}-3 \mathrm{t}^{2}$. Find the ratio of initial velocity and initial acceleration. \((\mathrm{A})+(5 / 6)\) (B) \(-(5 / 6)\) (C) \((6 / 5)\) (D) \(-(6 / 5)\)

$\quad\left(\mathrm{A}^{\rightarrow}+\mathrm{B}^{\rightarrow}\right) \cdot\left(\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\right)=\ldots$ (A) 0 (B) \(A^{2}+B^{2}\) (C) \(\sqrt{\left(A^{2}+B^{2}\right)}\) (D) \(\mathrm{A}^{2} \mathrm{~B}^{2}\)

If \(\mathrm{f}\) is the frequency of a body moving in a circular path with constant speed. a is its centrifugal acceleration, so. (A) \(\mathrm{a} \propto \mathrm{f}\) (B) a \(\propto \mathrm{f}^{2}\) (C) \(\mathrm{a} \propto \mathrm{f}^{3}\) (D) a \(\propto(1 / \mathrm{f})\)

To locate the position of the particle we need ... (A) a frame of reference (B) direction of the particle (C) size of the particle (D) mass of the particle
See all solutions

Recommended explanations on English Textbooks

Blog

Read Explanation

English Grammar

Read Explanation

Essay Prompts

Read Explanation

Essay Writing Skills

Read Explanation

Global English

Read Explanation

Creative Story

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