
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
233.1k+ 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
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

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

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

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

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

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

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

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

