
A ball starts rolling in a horizontal surface with an initial velocity $1\,\,m{s^{ - 1}}$. Due to friction its velocity decreases at the rate of $0.1\,\,m{s^{ - 2}}$.how much time will it take for the ball to stop?
A) $1\,\,s$ .
B) $100\,\,s$
C) $10\,\,s$.
D) $0.1\,\,s$.
Answer
124.2k+ views
Hint: For solving the above equation the formula for equation of motion can be used. There are two formulas for the equation of motion. They are non-accelerated motion and the accelerated motion. In case of the accelerated motion the object experiences varied velocity and acceleration. Since the ball with an initial velocity gradually decreases its speed and finally stops, we use the formula of the equation of accelerated motion.
Useful formula:
The equation of accelerated motion;
$v = u + at$
Where $v$ denotes the final velocity of the ball, $u$ denotes the initial velocity of the ball, $a$ denotes the acceleration of the ball, $t$ denotes the time taken by the ball.
Complete step by step solution:
Given data:
Initial velocity of the ball is $u = 1\,\,m{s^{ - 1}}$ ,
The acceleration of the ball is $a = 0.1\,\,m{s^{ - 2}}$ .
By using the equation for accelerated motion;
$v = u + at$
Substitute the values of initial velocity $u$ and the acceleration $a$ of the ball in the above equation;
Since the ball stops after a certain amount of time the final velocity is $v = 0$.
$0\,\,m{s^{ - 1}} = 1\,\,m{s^{ - 1}} + \left( {0.1\,\,m{s^{ - 2}} \times t} \right)$
On simplifying the equation, we get;
$0 = 1 + 0.1\,t$
$0.1\,t = 1$
Since we only need time taken by the ball;
$t = \dfrac{1}{{0.1}}$
$t = 10\,\,s$
Therefore, the time taken by the ball to stop moving is $t = 10\,\,s$ .
Hence, the option (C), $t = 10\,\,s$ is the correct answer.
Note: In case of non-accelerated motion, the object experiencing the motion is at constant velocity. There is no change in velocity. The equation of accelerated motion is derived from the basic formula that is acceleration is equal to the change in velocity to the time taken.
Useful formula:
The equation of accelerated motion;
$v = u + at$
Where $v$ denotes the final velocity of the ball, $u$ denotes the initial velocity of the ball, $a$ denotes the acceleration of the ball, $t$ denotes the time taken by the ball.
Complete step by step solution:
Given data:
Initial velocity of the ball is $u = 1\,\,m{s^{ - 1}}$ ,
The acceleration of the ball is $a = 0.1\,\,m{s^{ - 2}}$ .
By using the equation for accelerated motion;
$v = u + at$
Substitute the values of initial velocity $u$ and the acceleration $a$ of the ball in the above equation;
Since the ball stops after a certain amount of time the final velocity is $v = 0$.
$0\,\,m{s^{ - 1}} = 1\,\,m{s^{ - 1}} + \left( {0.1\,\,m{s^{ - 2}} \times t} \right)$
On simplifying the equation, we get;
$0 = 1 + 0.1\,t$
$0.1\,t = 1$
Since we only need time taken by the ball;
$t = \dfrac{1}{{0.1}}$
$t = 10\,\,s$
Therefore, the time taken by the ball to stop moving is $t = 10\,\,s$ .
Hence, the option (C), $t = 10\,\,s$ is the correct answer.
Note: In case of non-accelerated motion, the object experiencing the motion is at constant velocity. There is no change in velocity. The equation of accelerated motion is derived from the basic formula that is acceleration is equal to the change in velocity to the time taken.
Recently Updated Pages
JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main 2025 Helpline Numbers - Center Contact, Phone Number, Address

JEE Main Course 2025 - Important Updates and Details

NTA JEE Mains 2025 Correction window - Dates and Procedure

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

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

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

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

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main Login 2045: Step-by-Step Instructions and Details

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

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

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

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation
