A particle moves in a straight line, according to the law \[x=4a[t+a\sin (\dfrac{t}{a})]\], where \[x\] is its position in meters, \[t\] in sec. & \[a\] is some constants, then velocity is zero at
A) \[x=4{{a}^{2}}\pi meters\]
B) \[t=\pi \sec .\]
C) \[t=0\sec .\]
D) \[none\]
Answer
273.3k+ views
Hint: Velocity is the rate of change of displacement. To get velocity the equation of displacement must be differentiated with respect to time. Then solve as per the given question.
Complete step by step answer:
The given equation is for displacement and we are required to find the time when velocity is zero.
Since velocity is the rate of change of displacement hence we will differentiate the given equation.
$\Rightarrow x=4a[t+a\sin (\dfrac{t}{a})]$
Here \[x\] is the displacement
\[a\] is a constant
\[t\] is the time in seconds
Let’s first simplify this equation,
\[\Rightarrow x=4at+4{{a}^{2}}\sin (\dfrac{t}{a})\]
Now differentiating this equation with respect to time we will get equation of velocity since $v=\dfrac{dx}{dt}$
\[\Rightarrow \dfrac{dx}{dt}=4a+4{{a}^{2}}(\cos (\dfrac{t}{a}))(\dfrac{1}{a})\]
Here \[\dfrac{d(4at)}{dt}=4a\] since \[a\] is constant as given in the question.
And it is also known from basic differentiation rules that \[\dfrac{d(\sin \theta )}{d\theta }=\cos \theta \]
Also \[\dfrac{d(\sin a\theta )}{d\theta }=(\cos a\theta )(\dfrac{da\theta }{d\theta })\]
We finally get
\[\Rightarrow v=\dfrac{dx}{dt}=4a+4a\cos (\dfrac{t}{a})\]
That is equation of velocity is given as
\[\Rightarrow v=4a+4a\cos (\dfrac{t}{a})\]
We need to calculate time for which the velocity is zero, hence we will put \[v=0\] in the above equation
$\Rightarrow 0=4a+4a\cos (\dfrac{t}{a})$
On solving further we get,
$\Rightarrow 0=4a[1+\cos \dfrac{t}{a})]$
$\Rightarrow 0=1+\cos (\dfrac{t}{a})$
$\Rightarrow -1=\cos (\dfrac{t}{a})$
Now this equation is of the form, $\cos \theta =-1$
We need to find $\theta $ such that $\cos \theta =-1$, let’s check the graph of $\cos \theta $
The value of $\cos \theta $ is equal to $-1$ when the value of $\theta $ is $\pi $.
Therefore, $\cos (\dfrac{t}{a})=-1$
$\Rightarrow \dfrac{t}{a}=\pi $
$\Rightarrow t=a\pi $
Therefore the velocity is zero at $t=a\pi $ which is not given in the option.
Hence the correct option is $(D)$ none of these
Note: If we again differentiate the equation of velocity with respect to time, we will get an equation for acceleration. Do remember that there can be infinitely many solutions as the graph of \[\cos \theta \] is periodic in nature. It repeats after an interval of \[2\pi \]. This implies that the velocity will again be zero for \[t=\pi a+2\pi a\], \[t=\pi a+4\pi a\] and so on.
Complete step by step answer:
The given equation is for displacement and we are required to find the time when velocity is zero.
Since velocity is the rate of change of displacement hence we will differentiate the given equation.
$\Rightarrow x=4a[t+a\sin (\dfrac{t}{a})]$
Here \[x\] is the displacement
\[a\] is a constant
\[t\] is the time in seconds
Let’s first simplify this equation,
\[\Rightarrow x=4at+4{{a}^{2}}\sin (\dfrac{t}{a})\]
Now differentiating this equation with respect to time we will get equation of velocity since $v=\dfrac{dx}{dt}$
\[\Rightarrow \dfrac{dx}{dt}=4a+4{{a}^{2}}(\cos (\dfrac{t}{a}))(\dfrac{1}{a})\]
Here \[\dfrac{d(4at)}{dt}=4a\] since \[a\] is constant as given in the question.
And it is also known from basic differentiation rules that \[\dfrac{d(\sin \theta )}{d\theta }=\cos \theta \]
Also \[\dfrac{d(\sin a\theta )}{d\theta }=(\cos a\theta )(\dfrac{da\theta }{d\theta })\]
We finally get
\[\Rightarrow v=\dfrac{dx}{dt}=4a+4a\cos (\dfrac{t}{a})\]
That is equation of velocity is given as
\[\Rightarrow v=4a+4a\cos (\dfrac{t}{a})\]
We need to calculate time for which the velocity is zero, hence we will put \[v=0\] in the above equation
$\Rightarrow 0=4a+4a\cos (\dfrac{t}{a})$
On solving further we get,
$\Rightarrow 0=4a[1+\cos \dfrac{t}{a})]$
$\Rightarrow 0=1+\cos (\dfrac{t}{a})$
$\Rightarrow -1=\cos (\dfrac{t}{a})$
Now this equation is of the form, $\cos \theta =-1$
We need to find $\theta $ such that $\cos \theta =-1$, let’s check the graph of $\cos \theta $
The value of $\cos \theta $ is equal to $-1$ when the value of $\theta $ is $\pi $.
Therefore, $\cos (\dfrac{t}{a})=-1$
$\Rightarrow \dfrac{t}{a}=\pi $
$\Rightarrow t=a\pi $
Therefore the velocity is zero at $t=a\pi $ which is not given in the option.
Hence the correct option is $(D)$ none of these
Note: If we again differentiate the equation of velocity with respect to time, we will get an equation for acceleration. Do remember that there can be infinitely many solutions as the graph of \[\cos \theta \] is periodic in nature. It repeats after an interval of \[2\pi \]. This implies that the velocity will again be zero for \[t=\pi a+2\pi a\], \[t=\pi a+4\pi a\] and so on.
Recently Updated Pages
WBJEE 2026 Result Live: Important Dates, Last Date Apply Online 2026

JoSAA Counselling 2026: JoSAA 2026 Mock Seat Allotment LIVE: Round 2 Result Released, Registration, Choice Filling and Ranks

Circuit Switching vs Packet Switching: Key Differences Explained

Dimensions of Pressure in Physics: Formula, Derivation & SI Unit

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

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

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Derivation of Equation of Trajectory Explained for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2026-27

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

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2026-27

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

