
A particle initially (i.e., at $t = 0$) moving with a velocity u is subjected to a retarding force, as a result of which it decelerates $a = - k\sqrt v $ at a rate where v is the instantaneous velocity and k is a positive constant. The time T taken by the particle to come to rest is given by :
(A) $T = \dfrac{{2\sqrt u }}{k}$
(B) $T = \dfrac{{2u}}{k}$
(C) $T = \dfrac{{2{u^{\dfrac{3}{2}}}}}{k}$
(D) $T = \dfrac{{2{u^2}}}{k}$
Answer
163.2k+ views
Hint:In order to solve this question, we will use the derivative form of acceleration in terms of velocity and time and then using integration formulas we will find a relation between velocity and time and finally we will solve for time at which final velocity of the body becomes zero.
Formula used:
Acceleration ‘a’ can be written as $a = \dfrac{{dv}}{{dt}}$ where, ‘dv’ is the change in velocity an ‘dt’ is the change in time
Complete answer:
According to the question, we have given that at $t = 0$ initial velocity of the particle is ‘u’ and acceleration is given to us as $a = - k\sqrt v $ so, we can write acceleration in derivative form as $a = \dfrac{{dv}}{{dt}}$ so, we get
$\dfrac{{dv}}{{dt}} = - k\sqrt v $
Now, integrating both sides we get with the limits of velocity as $
{v_1} = u,{t_1} = 0 \\
{v_2} = v,{t_2} = t \\
$
$\int\limits_u^v {{v^{ - \dfrac{1}{2}}}} dv = - k\int\limits_0^t {dt} $
Using the integration formula $\int {{r^n}dr} = \dfrac{{{r^{n + 1}}}}{{n + 1}}$ we get,
$
[2\sqrt v ]_u^v = - k[t]_0^t \\
2[\sqrt v - \sqrt u ] = - kt \\
$
Now, we need to find at what time ‘t’ the final velocity of the particle becomes zero which means it became at rest, so putting the value of $v = 0$ and at this time denoting time t by ‘T’ we get,
$2[\sqrt 0 - \sqrt u ] = - kT$
solving for T we get,
$T = \dfrac{{2\sqrt u }}{k}$
So, after the time $T = \dfrac{{2\sqrt u }}{k}$ the particle will finally come to rest.
Hence, the correct option is Option (A).
Note:While solving such questions, involving differentiation and integration, always remember the basic integration formulas and derivative forms of physical quantities and avoid the integration constant using given limits of the variable, also remember constant terms do not come inside integrals as they are not variables.
Formula used:
Acceleration ‘a’ can be written as $a = \dfrac{{dv}}{{dt}}$ where, ‘dv’ is the change in velocity an ‘dt’ is the change in time
Complete answer:
According to the question, we have given that at $t = 0$ initial velocity of the particle is ‘u’ and acceleration is given to us as $a = - k\sqrt v $ so, we can write acceleration in derivative form as $a = \dfrac{{dv}}{{dt}}$ so, we get
$\dfrac{{dv}}{{dt}} = - k\sqrt v $
Now, integrating both sides we get with the limits of velocity as $
{v_1} = u,{t_1} = 0 \\
{v_2} = v,{t_2} = t \\
$
$\int\limits_u^v {{v^{ - \dfrac{1}{2}}}} dv = - k\int\limits_0^t {dt} $
Using the integration formula $\int {{r^n}dr} = \dfrac{{{r^{n + 1}}}}{{n + 1}}$ we get,
$
[2\sqrt v ]_u^v = - k[t]_0^t \\
2[\sqrt v - \sqrt u ] = - kt \\
$
Now, we need to find at what time ‘t’ the final velocity of the particle becomes zero which means it became at rest, so putting the value of $v = 0$ and at this time denoting time t by ‘T’ we get,
$2[\sqrt 0 - \sqrt u ] = - kT$
solving for T we get,
$T = \dfrac{{2\sqrt u }}{k}$
So, after the time $T = \dfrac{{2\sqrt u }}{k}$ the particle will finally come to rest.
Hence, the correct option is Option (A).
Note:While solving such questions, involving differentiation and integration, always remember the basic integration formulas and derivative forms of physical quantities and avoid the integration constant using given limits of the variable, also remember constant terms do not come inside integrals as they are not variables.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

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

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

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

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

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

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

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

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

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
