
The mosquito net over a \[7\;{\text{ft}} \times 4\;{\text{ft}}\]bed is $3\;{\text{ft}}$ high. The net has a hole at one corner of the bed through which a mosquito enters the net. Find the magnitude of the displacement of the mosquito.
A) $\sqrt {54} \;{\text{ft}}$
B) $81\;{\text{ft}}$
C) $\sqrt {74} \;{\text{ft}}$
D) $\sqrt {64} \;{\text{ft}}$
Answer
137.1k+ views
Hint: The above problem is based on the kinematics. The displacement is defined as the change in the position and direction of the object. The net over the bed is the same as the box in the shape of a cube. The displacement of the mosquito can be found by the formula to find the length of the diagonal of the box.
Complete step by step answer
Given: The length of the bed is $l = 7\;{\text{ft}}$, width of the bed is $b = 4\;{\text{ft}}$ and height of the net is $h = 3\;{\text{ft}}$.
The formula to calculate the magnitude of the displacement of the mosquito is given as:
$d = \sqrt {{l^2} + {b^2} + {h^2}} $
Substitute $7\;{\text{ft}}$for l, $4\;{\text{ft}}$for b and $3\;{\text{ft}}$for h in the above formula to find the magnitude of the displacement of the mosquito.
$d = \sqrt {{{\left( {7\;{\text{ft}}} \right)}^2} + {{\left( {4\;{\text{ft}}} \right)}^2} + {{\left( {3\;{\text{ft}}} \right)}^2}} $
$d = \sqrt {74} \;{\text{ft}}$
Thus, the magnitude of the displacement of the mosquito is $\sqrt {74} \;{\text{ft}}$and the
option (A) is the correct answer.
Additional Information: The displacement and distance of the object is different. The displacement consists both magnitude and direction while distance only consists of magnitude. The change in the displacement of the object with time is the same as the velocity and change in the distance of the object with time is the same as the speed of the object.
Note: The displacement of the mosquito can also be found by using the vector method. The one corner of the net assumed as the origin and coordinates as (0, 0, 0) and the coordinates of the mosquito as (7 ft, 4 ft, 3 ft). Then find the distance between these two coordinates to find the magnitude of the mosquito.
Complete step by step answer
Given: The length of the bed is $l = 7\;{\text{ft}}$, width of the bed is $b = 4\;{\text{ft}}$ and height of the net is $h = 3\;{\text{ft}}$.
The formula to calculate the magnitude of the displacement of the mosquito is given as:
$d = \sqrt {{l^2} + {b^2} + {h^2}} $
Substitute $7\;{\text{ft}}$for l, $4\;{\text{ft}}$for b and $3\;{\text{ft}}$for h in the above formula to find the magnitude of the displacement of the mosquito.
$d = \sqrt {{{\left( {7\;{\text{ft}}} \right)}^2} + {{\left( {4\;{\text{ft}}} \right)}^2} + {{\left( {3\;{\text{ft}}} \right)}^2}} $
$d = \sqrt {74} \;{\text{ft}}$
Thus, the magnitude of the displacement of the mosquito is $\sqrt {74} \;{\text{ft}}$and the
option (A) is the correct answer.
Additional Information: The displacement and distance of the object is different. The displacement consists both magnitude and direction while distance only consists of magnitude. The change in the displacement of the object with time is the same as the velocity and change in the distance of the object with time is the same as the speed of the object.
Note: The displacement of the mosquito can also be found by using the vector method. The one corner of the net assumed as the origin and coordinates as (0, 0, 0) and the coordinates of the mosquito as (7 ft, 4 ft, 3 ft). Then find the distance between these two coordinates to find the magnitude of the mosquito.
Recently Updated Pages
COM of Semicircular Ring Important Concepts and Tips for JEE

Geostationary Satellites and Geosynchronous Satellites for JEE

Current Loop as Magnetic Dipole Important Concepts for JEE

Electromagnetic Waves Chapter for JEE Main Physics

Structure of Atom: Key Models, Subatomic Particles, and Quantum Numbers

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

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

A body crosses the topmost point of a vertical circle class 11 physics JEE_Main

JEE Main 2025: Derivation of Equation of Trajectory in Physics

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

Degree of Dissociation and Its Formula With Solved Example for JEE

At which height is gravity zero class 11 physics JEE_Main

Other Pages
Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

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

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

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

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

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