
Two cars A and B moving with velocity $36 km/h$ and $54km/h$ moving along a straight line the car B follow car A. The length of each car is $5m$ . The time taken by car B to cross car A and also find distance travelled by both cars during this time?
Answer
233.4k+ views
Hint: For solving this question we have to calculate the time when car B crosses car A. And we have to calculate the distance at the crossing time which is the final speed $ \times $ time. For calculating the time, we know distance upon speed which means length of car A $ + $ car B and then divide by relative velocity. For that we have different velocity for both the car and for that we have to find out first relative velocity. Knowing all these values and after substituting we can get the required answer.
Complete step by step solution:
Distance between the end of car B and the front point of car A is $10m$.
Now,
${v_{relative}} = 54 - 36$
We get $18km/h $
Now, we have to calculate time taken to travel 10m is $t = \dfrac{{10 \times 10}}{{18}}$
Now converting kilometres to meters, we get $t = \dfrac{5}{9} \times {10^{ - 3}}$
Now, we have to find out the total distance travelled by car B over car A.
That is, $t = \dfrac{5}{9} \times {10^{ - 3}} \times 54 \times 1000$
$\therefore t = 30m$
So, the time taken by car b to cross car A is $30m$.
So, The correct answer is $30m$.
Note: For solving this question we have to find out the total time taken by car B while crossing Car A. For solving this question we have to calculate the distance between both cars and how much take it takes. Since, we know distance can be defined as the length of the space between two points this is known as distance. We have to calculate relative velocity. By substituting all these values, we can get the correct answer. So, these are some important points.
Complete step by step solution:
Distance between the end of car B and the front point of car A is $10m$.
Now,
${v_{relative}} = 54 - 36$
We get $18km/h $
Now, we have to calculate time taken to travel 10m is $t = \dfrac{{10 \times 10}}{{18}}$
Now converting kilometres to meters, we get $t = \dfrac{5}{9} \times {10^{ - 3}}$
Now, we have to find out the total distance travelled by car B over car A.
That is, $t = \dfrac{5}{9} \times {10^{ - 3}} \times 54 \times 1000$
$\therefore t = 30m$
So, the time taken by car b to cross car A is $30m$.
So, The correct answer is $30m$.
Note: For solving this question we have to find out the total time taken by car B while crossing Car A. For solving this question we have to calculate the distance between both cars and how much take it takes. Since, we know distance can be defined as the length of the space between two points this is known as distance. We have to calculate relative velocity. By substituting all these values, we can get the correct answer. So, these are some important points.
Recently Updated Pages
JEE Main 2026 Session 2 Registration Open, Exam Dates, Syllabus & Eligibility

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

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

Understanding Average and RMS Value in Electrical Circuits

Understanding Collisions: Types and Examples for Students

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Understanding Atomic Structure for Beginners

Derive an expression for maximum speed of a car on class 11 physics JEE_Main

Other Pages
JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions For Class 11 Physics Chapter 9 Mechanical Properties of Fluids (2025-26)

NCERT Solutions For Class 11 Physics Chapter 12 Kinetic Theory (2025-26)

NCERT Solutions For Class 11 Physics Chapter 4 Law of Motion (2025-26)

Class 11 JEE Main Physics Mock Test 2025

Inductive Effect and Its Role in Acidic Strength

