
A particle moves along a straight line OX. At a time t (in seconds) the distance x (in meters) of the particle from 0 is given by: \[x = 40 + 12t - {t^3}\]. How long would the particles travel before coming to rest?
A) 24m
B) 40m
C) 56m
D) 16m
Answer
138k+ views
Hint: The distance is defined as the length of the path between the initial position and the final position of a moving object in the given interval of time. It is a scalar quantity. The change in the position of an object is called displacement. Displacement is a vector quantity and has both magnitude and direction.
Complete step by step solution:
Given equation \[ x = 40 + 12t - t^3\].....................(1)
Let say that the particle is at a distance x from O at a time, t=0
Thus substituting the value of t=0 in the given equation, \[x = 40 + 12t - {t^3}\], we get,
\[x = 40 + 12\left( 0 \right) - {\left( 0 \right)^3} = 40m\]
It is given that the particle will come to rest which means the velocity of the particle becomes zero after traveling to a certain displacement and let us consider this time as t.
Differentiating the given equation 1 to time, we get,
$\Rightarrow v = 12 - 3{t^2}$.................(2)
When the particle comes to rest at time=t, velocity=0
Substituting the value of v=0 in equation 2, we get,
$\Rightarrow 12 - 3{t^2} = 0$
$ \Rightarrow 3{t^2} = 12$
$ \Rightarrow {t^2} = 4$
$ \Rightarrow t = 2s$
Substituting the value of t in equation 1, we get
\[\Rightarrow {x} = 40 + 12\left( 2 \right) - {\left( 2 \right)^3}\]
\[ \Rightarrow {x} = 56m\]
Thus, it is clear that the particle had started the journey when it was at a distance of 40m from point O and came to rest at 56m from point O.
Thus, the particle had traveled a distance of $56m - 40m = 16m$.
Thus, the particle had traveled a distance of 16m before coming to rest.
Hence the correct option is D.
Note: 1. The velocity is defined as the rate of change of the position of the object to time. Velocity is also a vector quantity and has both direction and magnitude.
2. The distance and the displacement have the same SI unit, meter.
Complete step by step solution:
Given equation \[ x = 40 + 12t - t^3\].....................(1)
Let say that the particle is at a distance x from O at a time, t=0
Thus substituting the value of t=0 in the given equation, \[x = 40 + 12t - {t^3}\], we get,
\[x = 40 + 12\left( 0 \right) - {\left( 0 \right)^3} = 40m\]
It is given that the particle will come to rest which means the velocity of the particle becomes zero after traveling to a certain displacement and let us consider this time as t.
Differentiating the given equation 1 to time, we get,
$\Rightarrow v = 12 - 3{t^2}$.................(2)
When the particle comes to rest at time=t, velocity=0
Substituting the value of v=0 in equation 2, we get,
$\Rightarrow 12 - 3{t^2} = 0$
$ \Rightarrow 3{t^2} = 12$
$ \Rightarrow {t^2} = 4$
$ \Rightarrow t = 2s$
Substituting the value of t in equation 1, we get
\[\Rightarrow {x} = 40 + 12\left( 2 \right) - {\left( 2 \right)^3}\]
\[ \Rightarrow {x} = 56m\]
Thus, it is clear that the particle had started the journey when it was at a distance of 40m from point O and came to rest at 56m from point O.
Thus, the particle had traveled a distance of $56m - 40m = 16m$.
Thus, the particle had traveled a distance of 16m before coming to rest.
Hence the correct option is D.
Note: 1. The velocity is defined as the rate of change of the position of the object to time. Velocity is also a vector quantity and has both direction and magnitude.
2. The distance and the displacement have the same SI unit, meter.
Recently Updated Pages
COM of Semicircular Ring Important Concepts and Tips for JEE

Geostationary Satellites and Geosynchronous Satellites for JEE

Current Loop as Magnetic Dipole Important Concepts for JEE

Electromagnetic Waves Chapter for JEE Main Physics

Structure of Atom: Key Models, Subatomic Particles, and Quantum Numbers

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

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

A body crosses the topmost point of a vertical circle class 11 physics JEE_Main

JEE Main 2025: Derivation of Equation of Trajectory in Physics

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

Degree of Dissociation and Its Formula With Solved Example for JEE

At which height is gravity zero class 11 physics JEE_Main

Other Pages
Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

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

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

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

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

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