
A man travelling in a car with a maximum constant speed of \[20\text{ }ms{{~}^{-1}}\] watches the friend start off at a distance 100 m ahead on a motorcycle with constant acceleration ‘a’. The maximum value of ‘a’ for which the man in the car can reach his friend is ?

A. $2\,m{{s}^{-1}}$
B. $1\,m{{s}^{-1}}$
C. $4\,m{{s}^{-1}}$
D. None of these
Answer
163.8k+ views
Hint:To do this question you have to know about Newton’s equation of motion. By applying the third equation of motion, we can solve this question. Here we have to find the maximum acceleration attained by the car to reach the bike. So, extreme conditions are also applied.
Formula used:
Here Newton's third equation of motion is used. Let a be the acceleration and u be the initial velocity and v be the final velocity and s be the distance travelled by the particle. Then, newton’s third equation of motion is:
${{v}^{2}}={{u}^{2}}+2as$
Complete step by step solution:
The car is travelling with a maximum constant speed of \[20\text{ }ms{{~}^{-1}}\]. Motorcycle which is 100 m away from the car is just starting and it is moving with constant acceleration. We have to find the acceleration attained by the car to reach the motorcycle.
If the man has to reach his friend, then his friend should not be faster than him until he covers the distance of 100 m. From the question it is clear that the initial velocity of the motorcycle is zero. That is, on applying Newton’s third equation of motion we get:
${{20}^{2}}\ge 200a$
That is,
$a<2$
From this condition we can say that the maximum acceleration of a car should be $2\,m{{s}^{-1}}$ to reach a motorcycle which is 100 m away from the car.
Hence, the correct answer is option A.
Note: If we know the basics, then this is a simple question. In this question without applying extreme conditions, we can find maximum acceleration attained by the car to reach the motor cycle which is 100 m away from the car.
Formula used:
Here Newton's third equation of motion is used. Let a be the acceleration and u be the initial velocity and v be the final velocity and s be the distance travelled by the particle. Then, newton’s third equation of motion is:
${{v}^{2}}={{u}^{2}}+2as$
Complete step by step solution:
The car is travelling with a maximum constant speed of \[20\text{ }ms{{~}^{-1}}\]. Motorcycle which is 100 m away from the car is just starting and it is moving with constant acceleration. We have to find the acceleration attained by the car to reach the motorcycle.
If the man has to reach his friend, then his friend should not be faster than him until he covers the distance of 100 m. From the question it is clear that the initial velocity of the motorcycle is zero. That is, on applying Newton’s third equation of motion we get:
${{20}^{2}}\ge 200a$
That is,
$a<2$
From this condition we can say that the maximum acceleration of a car should be $2\,m{{s}^{-1}}$ to reach a motorcycle which is 100 m away from the car.
Hence, the correct answer is option A.
Note: If we know the basics, then this is a simple question. In this question without applying extreme conditions, we can find maximum acceleration attained by the car to reach the motor cycle which is 100 m away from the car.
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
