
An object, moving with a speed of $6.25\,m{s^{ - 1}}$ , is decelerated at a rate given by: $\dfrac{{dv}}{{dt}} = - 2.5\sqrt v $ where $v$ is the instantaneous speed. The time taken by the object, to come to rest, would be:
A) $2s$
B) $4s$
C) $8s$
D) $1s$
Answer
216.6k+ views
Hint: The deceleration rate is given in terms of the differential equation; we need to calculate the time taken for the body to come to rest. Integrate the equation to get the instantaneous speed in terms of time and then put the value of initial speed as $6.25\,m{s^{ - 1}}$ to find the time taken.
Complete step by step solution:
We are given with the initial speed, $u = 6.25\,m{s^{ - 1}}$
The rate of deceleration is given as:
$\dfrac{{dv}}{{dt}} = - 2.5\sqrt v $
The negative sign indicates that the body is decelerating.
$ \Rightarrow \dfrac{{dv}}{{\sqrt v }} = - 2.5dt$
Integrating the above equation, we get
$ \Rightarrow \int\limits_{6.25}^0 {\dfrac{{dv}}{{\sqrt v }}} = - 2.5\int\limits_0^t {dt} $
$ \therefore \left| {2\sqrt v } \right|_{6.25}^0 = - 2.5t$
Let the time taken be $t$ seconds and as the body comes to rest therefore, the final velocity will be zero.
Substituting the limits, we get:
$ \Rightarrow 2\sqrt {6.25} = - 2.5t$
$ \therefore t = 2s$
Therefore, the time taken by the object to come to rest is $2$ seconds.
Option A is the correct option.
Additional details: The acceleration of a body is the rate of change in velocity and it can be expressed as $\dfrac{{dv}}{{dt}}$ . Similarly, as velocity is rate of change of displacement, it can also be represented as $\dfrac{{dx}}{{dt}}$. Deceleration is the negative of acceleration. For a body which is decelerating, it will eventually come to rest at some point in time. Instantaneous speed is the speed of an object at a particular moment in time. If we include the direction with that speed, we get the instantaneous velocity.
Note: Be careful with the limits after integrating. Remember that as the body comes to rest eventually hence, its final speed must be zero. Also remember that the negative sign indicates that the body is decelerating. As time cannot be negative hence, we have taken only positive values of time.
Complete step by step solution:
We are given with the initial speed, $u = 6.25\,m{s^{ - 1}}$
The rate of deceleration is given as:
$\dfrac{{dv}}{{dt}} = - 2.5\sqrt v $
The negative sign indicates that the body is decelerating.
$ \Rightarrow \dfrac{{dv}}{{\sqrt v }} = - 2.5dt$
Integrating the above equation, we get
$ \Rightarrow \int\limits_{6.25}^0 {\dfrac{{dv}}{{\sqrt v }}} = - 2.5\int\limits_0^t {dt} $
$ \therefore \left| {2\sqrt v } \right|_{6.25}^0 = - 2.5t$
Let the time taken be $t$ seconds and as the body comes to rest therefore, the final velocity will be zero.
Substituting the limits, we get:
$ \Rightarrow 2\sqrt {6.25} = - 2.5t$
$ \therefore t = 2s$
Therefore, the time taken by the object to come to rest is $2$ seconds.
Option A is the correct option.
Additional details: The acceleration of a body is the rate of change in velocity and it can be expressed as $\dfrac{{dv}}{{dt}}$ . Similarly, as velocity is rate of change of displacement, it can also be represented as $\dfrac{{dx}}{{dt}}$. Deceleration is the negative of acceleration. For a body which is decelerating, it will eventually come to rest at some point in time. Instantaneous speed is the speed of an object at a particular moment in time. If we include the direction with that speed, we get the instantaneous velocity.
Note: Be careful with the limits after integrating. Remember that as the body comes to rest eventually hence, its final speed must be zero. Also remember that the negative sign indicates that the body is decelerating. As time cannot be negative hence, we have taken only positive values of time.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Electricity and Magnetism Explained: Key Concepts & Applications

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

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

NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation 2025-26

Understanding Atomic Structure for Beginners

