
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
145.2k+ 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
Difference Between Vapor and Gas: JEE Main 2024

Area of an Octagon Formula - Explanation, and FAQs

Charle's Law Formula - Definition, Derivation and Solved Examples

Central Angle of a Circle Formula - Definition, Theorem and FAQs

Average Force Formula - Magnitude, Solved Examples and FAQs

Boyles Law Formula - Boyles Law Equation | Examples & Definitions

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

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

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry
