A particle is moving along the x-axis whose instantaneous speed is given by ${v^2} = 108 - 9{x^2}$. The acceleration of the particle is:
(A) $- \dfrac{{9x}}{2}m{s^{ - 2}}$
(B) $- 18xm{s^{ - 2}}$
(C) $- \dfrac{{9x}}{2}m{s^{ - 2}}$
(D) None of these
Answer
249.6k+ views
Hint We know that when the speed of an object is constantly changing, the instantaneous speed is the speed of an object at a particular moment or we can say instant in time. Based on this concept we have to solve this question, by derivation of the given equation.
Complete step by step answer
We know that it is given:
${v^2} = 108 - 9{x^2}$
So, we can write this as:
$\dfrac{{dv}}{{dx}}\dfrac{{dx}}{{dt}} = v\dfrac{{dv}}{{dx}}$
As we know a = $\dfrac{{dv}}{{dx}}\dfrac{{dx}}{{dt}} = v\dfrac{{dv}}{{dx}}$
So, the value of a = $- 18xm{s^{ - 2}}$.
Hence, option B is correct.
Note For solving such problems we know that the rate of change of distance of an object with the respect to time, is an idea that is obtained from the concept of instantaneous speed. The unit of this physical quantity is given as distance divided by time.
Complete step by step answer
We know that it is given:
${v^2} = 108 - 9{x^2}$
So, we can write this as:
$\dfrac{{dv}}{{dx}}\dfrac{{dx}}{{dt}} = v\dfrac{{dv}}{{dx}}$
As we know a = $\dfrac{{dv}}{{dx}}\dfrac{{dx}}{{dt}} = v\dfrac{{dv}}{{dx}}$
So, the value of a = $- 18xm{s^{ - 2}}$.
Hence, option B is correct.
Note For solving such problems we know that the rate of change of distance of an object with the respect to time, is an idea that is obtained from the concept of instantaneous speed. The unit of this physical quantity is given as distance divided by time.
Recently Updated Pages
JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Instantaneous Velocity Explained: Formula, Examples & Graphs

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Hybridisation in Chemistry – Concept, Types & Applications

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2025-26

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2025-26

CBSE Notes Class 11 Physics Chapter 4 - Laws of Motion - 2025-26

