
Two cars are travelling in the same direction with a velocity of $ 60\,{{{\rm{km}}} {\left/
{\vphantom {{{\rm{km}}} {\rm{h}}}} \right.} {\rm{h}}}$ . They are separated by a distance of $5\,{\rm{km}} $. A truck moving in the opposite direction meets the two cars in a time interval of $3\min$. The velocity of the truck is (in${{{\rm{km}}} {\left/ {\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}} $)
A. 20
B. 30
C. 40
D. 60
Answer
570.6k+ views
Hint: The velocity of a particle is defined as distance covered per unit time taken. When two bodies move in the opposite direction then their velocities add up to calculate the relative velocity and when two bodies are in the same direction then their velocities are subtracted to calculate the relative velocity.
Complete step by step answer:
Given: The velocity of two cars is$60\,{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {\rm{h}}}} \right.} {\rm{h}}}$. The distance between two cars and a truck is $5\,{\rm{km}}$. The time taken by the truck to meet cars is $3\,\min $.
The formula to calculate the relative velocity of the truck with respect to cars is ${v_t} = \dfrac{d}{t}$
Here, ${v_t}$ is the relative velocity of the truck, $d$ is the distance covered and $t$ is the time taken to cover the distance.
Substitute $5\,{\rm{km}}$ for$d$ and$3\,\min $ for $t$ in the formula to calculate the relative velocity of the truck.
$
{v_t} = \dfrac{{5\,{\rm{km}}}}{{\left( {3\,\min } \right)\left( {\dfrac{{1\,{\rm{h}}}}{{60\,\min }}} \right)}}\\
= \dfrac{{5\,{\rm{km}}}}{{0.05\,{\rm{h}}}}\\
= 100\,{{{\rm{km}}} {\left/
{\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}
$
Since the truck moves in the opposite direction of the cars, thus the formula to calculate the relative velocity of the truck is
${v_t} = v + {v_c}$
Here, $v$ is the velocity of a truck and ${v_c}$ is the velocity of a car.
Substitute $100\,{{{\rm{km}}} {\left/ {\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}$ for ${v_t}$ and $60\,{{{\rm{km}}} {\left/ {\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}$ for ${v_c}$ in the formula and solve to calculate the velocity of the truck.
$100\,{{{\rm{km}}} {\left/ {\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}} = v + 60{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}\\ $
$\implies v = 100{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}} -60{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}\\ $
$ = 40{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}$
Thus, the velocity of truck is $40{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {\rm{h}}}} \right.} {\rm{h}}}$
So, the correct answer is “Option C”.
Note:
The relative velocity of the system increases when the particles move in the opposite direction but it decreases when the particles move in the same direction.
Complete step by step answer:
Given: The velocity of two cars is$60\,{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {\rm{h}}}} \right.} {\rm{h}}}$. The distance between two cars and a truck is $5\,{\rm{km}}$. The time taken by the truck to meet cars is $3\,\min $.
The formula to calculate the relative velocity of the truck with respect to cars is ${v_t} = \dfrac{d}{t}$
Here, ${v_t}$ is the relative velocity of the truck, $d$ is the distance covered and $t$ is the time taken to cover the distance.
Substitute $5\,{\rm{km}}$ for$d$ and$3\,\min $ for $t$ in the formula to calculate the relative velocity of the truck.
$
{v_t} = \dfrac{{5\,{\rm{km}}}}{{\left( {3\,\min } \right)\left( {\dfrac{{1\,{\rm{h}}}}{{60\,\min }}} \right)}}\\
= \dfrac{{5\,{\rm{km}}}}{{0.05\,{\rm{h}}}}\\
= 100\,{{{\rm{km}}} {\left/
{\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}
$
Since the truck moves in the opposite direction of the cars, thus the formula to calculate the relative velocity of the truck is
${v_t} = v + {v_c}$
Here, $v$ is the velocity of a truck and ${v_c}$ is the velocity of a car.
Substitute $100\,{{{\rm{km}}} {\left/ {\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}$ for ${v_t}$ and $60\,{{{\rm{km}}} {\left/ {\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}$ for ${v_c}$ in the formula and solve to calculate the velocity of the truck.
$100\,{{{\rm{km}}} {\left/ {\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}} = v + 60{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}\\ $
$\implies v = 100{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}} -60{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}\\ $
$ = 40{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}$
Thus, the velocity of truck is $40{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {\rm{h}}}} \right.} {\rm{h}}}$
So, the correct answer is “Option C”.
Note:
The relative velocity of the system increases when the particles move in the opposite direction but it decreases when the particles move in the same direction.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

