
A car of mass 1000kg is moving at a speed of 30m/s. Brakes are applied to bring the car to rest. If the net retarding force is 5000N, the car comes to a stop after travelling ‘d’ m in t sec. then
A. d = 150, t = 5
B. d = 120, t = 8
C. d = 180, t = 6
D. d = 90, t = 6
Answer
164.7k+ views
Hint: In this question, we are given a car which is moving at a speed of 30 m/s. we have to find the distance and time at which the car stops when we apply the retarding force. First we find the acceleration by using the formula $f = ma$. Then we use the first and third equation of motion to find out the distance and time.
Formula used:
The formula of first equation of motion and third equation of motion :-
$v = u + at$
And, ${{v}^{2}}-{{u}^{2}}=2as$
Here, $v$ = Final velocity
$u$ = Initial velocity
$a$ = Acceleration
$t$ = time
Complete step by step solution:
Given a car of mass 1000 kg is moving at a speed of 30 m/s. And the net retarding force is 5000 N. We have to find out the car comes to a stop after travelling d m in t seconds.
We know retarding force $F = ma$
As the force applies is opposite to the motion, then;
$a = \dfrac{-5000}{1000}$= - 5 m/\[{{s}^{2}}\]
when the car stops, ${{v}_{\delta }}=0$
We know according to newton’s first equation of motion,
${{v}_{\delta }}-u=at$
Then $0 – 30 = -5t$
That is t = $\dfrac{-30}{-5}$= 6 sec
Now we know the third equation of motion,
${{v}^{2}}_{\delta }-{{u}^{2}}=2ad$
That is $0-{{30}^{2}}=-2\times 5d$
Then $d=\dfrac{-30\times 30}{-2\times 5}$= 90 m
It means the car comes to a stop after travelling 90 m in 6 seconds.
Thus, option D is correct.
Note: Retarding force is the force that resists relative motion. Sometimes they are also called just retarding forces. Whether the forces actually resist motion depends on who’s looking at a particular situation.
Formula used:
The formula of first equation of motion and third equation of motion :-
$v = u + at$
And, ${{v}^{2}}-{{u}^{2}}=2as$
Here, $v$ = Final velocity
$u$ = Initial velocity
$a$ = Acceleration
$t$ = time
Complete step by step solution:
Given a car of mass 1000 kg is moving at a speed of 30 m/s. And the net retarding force is 5000 N. We have to find out the car comes to a stop after travelling d m in t seconds.
We know retarding force $F = ma$
As the force applies is opposite to the motion, then;
$a = \dfrac{-5000}{1000}$= - 5 m/\[{{s}^{2}}\]
when the car stops, ${{v}_{\delta }}=0$
We know according to newton’s first equation of motion,
${{v}_{\delta }}-u=at$
Then $0 – 30 = -5t$
That is t = $\dfrac{-30}{-5}$= 6 sec
Now we know the third equation of motion,
${{v}^{2}}_{\delta }-{{u}^{2}}=2ad$
That is $0-{{30}^{2}}=-2\times 5d$
Then $d=\dfrac{-30\times 30}{-2\times 5}$= 90 m
It means the car comes to a stop after travelling 90 m in 6 seconds.
Thus, option D is correct.
Note: Retarding force is the force that resists relative motion. Sometimes they are also called just retarding forces. Whether the forces actually resist motion depends on who’s looking at a particular situation.
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

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

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

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

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

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

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

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