
A driver takes 0.20 s to apply the brakes after he sees a need for it. This is called the reaction time of the driver. If he is driving a car at the speed of 54km/h and the brakes cause a deceleration of $6.0m/{s^2}$, find the distance travelled by the car after he sees the need to put the brakes on.
$
A. 25 \\
B. 21.75 \\
C. 15 \\
D. 11.75 \\
$
Answer
224.7k+ views
Hint- Here we will proceed by using the formula of $Dis\tan ce = speed \times time$ and ${v^2} - {u^2} = 2as$ i.e. the Newton’s third law of motion. Then by substituting the values of initial velocity, distance and deceleration, we will get the required result.
Formulas used-
$Dis\tan ce = speed \times time$
${v^2} - {u^2} = 2as$( newton’s third law of motion)
Complete step-by-step answer:
We are given that-
Initial velocity is u = 54km/h
Time t = 0.2 sec
Final velocity is v = 0m/s
Deceleration is $d = 6.0m/{s^2}$
Now, we will calculate distance travelled by car during reaction time-
$Dis\tan ce = speed \times time$
Firstly, we will convert the initial velocity in km/h to m/s
u = 54km/h
or $u = 54 \times \dfrac{5}{{18}}$
or u = 15m/s (here u is distance)
Now, substituting the given values of speed i.e. 15m/s and time = 0.2 s
We get-
$Dis\tan ce = 15 \times 0.2$
So, the distance is 3m.
Using newton’s third law of motion i.e. For every action in nature, there is an equal and opposite reaction.
We get-
${v^2} - {u^2} = 2as$
Substituting the given values of v=0, u=15 in the third law of motion,
We get-
${0^2} - 15 \times 15 = 2*6*s$
$\Rightarrow s = \dfrac{{ - 15 \times 15}}{{ - 12}}$
$\Rightarrow s = \dfrac{{225}}{{12}}$
$\Rightarrow$ s = 18.75m
So, the total distance travelled by the driver after he sees that there is need to put the brakes on-
18.75 + 3 = 21.75m
Therefore, the option B is correct.
Note- In order to solve this type of question, we must know the three laws of motion as here we one of its laws i.e. for every action in nature, there is an equal and opposite reaction. Also we should make sure that we write SI units along with the answer.
Formulas used-
$Dis\tan ce = speed \times time$
${v^2} - {u^2} = 2as$( newton’s third law of motion)
Complete step-by-step answer:
We are given that-
Initial velocity is u = 54km/h
Time t = 0.2 sec
Final velocity is v = 0m/s
Deceleration is $d = 6.0m/{s^2}$
Now, we will calculate distance travelled by car during reaction time-
$Dis\tan ce = speed \times time$
Firstly, we will convert the initial velocity in km/h to m/s
u = 54km/h
or $u = 54 \times \dfrac{5}{{18}}$
or u = 15m/s (here u is distance)
Now, substituting the given values of speed i.e. 15m/s and time = 0.2 s
We get-
$Dis\tan ce = 15 \times 0.2$
So, the distance is 3m.
Using newton’s third law of motion i.e. For every action in nature, there is an equal and opposite reaction.
We get-
${v^2} - {u^2} = 2as$
Substituting the given values of v=0, u=15 in the third law of motion,
We get-
${0^2} - 15 \times 15 = 2*6*s$
$\Rightarrow s = \dfrac{{ - 15 \times 15}}{{ - 12}}$
$\Rightarrow s = \dfrac{{225}}{{12}}$
$\Rightarrow$ s = 18.75m
So, the total distance travelled by the driver after he sees that there is need to put the brakes on-
18.75 + 3 = 21.75m
Therefore, the option B is correct.
Note- In order to solve this type of question, we must know the three laws of motion as here we one of its laws i.e. for every action in nature, there is an equal and opposite reaction. Also we should make sure that we write SI units along with the answer.
Recently Updated Pages
Uniform Acceleration Explained: Formula, Examples & Graphs

JEE Main 2026 Session 1 Correction Window Started: Check Dates, Edit Link & Fees

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Trending doubts
JEE Main 2026: City Intimation Slip and Exam Dates Released, Application Form Closed, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

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

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

