
A ship of mass \[3 \times {10^7}kg\] initially at rest, is pulled by a force of \[5 \times {10^4}N\] through a distance of 3m. Assuming that the resistance due to water is negligible, the speed of the ship is:
A. 1.5 m/s
B. 60 m/s
C. 0.1 m/s
D. 5 m/s
Answer
233.1k+ views
Hint:At starting the ship is at rest so initial velocity becomes zero. The third equation of motion gives the relation between velocity and distance covered. By applying this we can easily find velocity acquired by a body in traveling a distance S. Also here we can use Newton’s second law which gives the relation between force, mass and acceleration.
Formula used
Newton’s second law,\[F = ma\]
Where, F is force, m is mass and a is acceleration.
Third equation of motion, \[{v^2} = {u^2} + 2aS\]
Where, v is final velocity, u is initial velocity, a is acceleration and S is distance.
Complete step by step solution:
Given: Mass of a ship, m=\[3 \times {10^7}kg\]
Force applied by ship, F=\[5 \times {10^4}N\]
Distance covered, S=3m
Initial velocity, u=0
According to Newton’s second law, we find an acceleration
\[F = ma\]
\[\Rightarrow a = \dfrac{F}{m}\]
\[\Rightarrow a = \dfrac{{5 \times {{10}^4}}}{{3 \times {{10}^7}}}\]
\[\Rightarrow a = \dfrac{5}{3} \times {10^{ - 3}}{\rm{m/se}}{{\rm{c}}^2}\]
Now by using this value we can find the speed of the ship after moving 3m.
By Third equation of motion, we get
\[{v^2} = {u^2} + 2aS\]
Putting the given value, we get
\[{v^2} = 0 + 2 \times \dfrac{5}{3} \times {10^{ - 3}} \times 3\]
\[\Rightarrow v = \sqrt {{{10}^{ - 2}}} {\rm{ }}\]
\[\Rightarrow v = {10^{ - 1}}\]
\[\therefore v = 0.1{\rm{m/sec}}\]
Therefore the speed of the ship is 0.1 m/sec.
Hence option C is the correct answer.
Note:The three equations of motion are related to some quantities which describe the motion of an object. The quantities are displacement S, initial velocity u, final velocity v, acceleration a, time taken t. Whether the object is at rest or in motion by using three equations of motion we obtain a relation between them. All quantities are linked with each other by three equations of motion. They are also known as kinematic equations of motion. The word kinematic means something which is in motion.
Formula used
Newton’s second law,\[F = ma\]
Where, F is force, m is mass and a is acceleration.
Third equation of motion, \[{v^2} = {u^2} + 2aS\]
Where, v is final velocity, u is initial velocity, a is acceleration and S is distance.
Complete step by step solution:
Given: Mass of a ship, m=\[3 \times {10^7}kg\]
Force applied by ship, F=\[5 \times {10^4}N\]
Distance covered, S=3m
Initial velocity, u=0
According to Newton’s second law, we find an acceleration
\[F = ma\]
\[\Rightarrow a = \dfrac{F}{m}\]
\[\Rightarrow a = \dfrac{{5 \times {{10}^4}}}{{3 \times {{10}^7}}}\]
\[\Rightarrow a = \dfrac{5}{3} \times {10^{ - 3}}{\rm{m/se}}{{\rm{c}}^2}\]
Now by using this value we can find the speed of the ship after moving 3m.
By Third equation of motion, we get
\[{v^2} = {u^2} + 2aS\]
Putting the given value, we get
\[{v^2} = 0 + 2 \times \dfrac{5}{3} \times {10^{ - 3}} \times 3\]
\[\Rightarrow v = \sqrt {{{10}^{ - 2}}} {\rm{ }}\]
\[\Rightarrow v = {10^{ - 1}}\]
\[\therefore v = 0.1{\rm{m/sec}}\]
Therefore the speed of the ship is 0.1 m/sec.
Hence option C is the correct answer.
Note:The three equations of motion are related to some quantities which describe the motion of an object. The quantities are displacement S, initial velocity u, final velocity v, acceleration a, time taken t. Whether the object is at rest or in motion by using three equations of motion we obtain a relation between them. All quantities are linked with each other by three equations of motion. They are also known as kinematic equations of motion. The word kinematic means something which is in motion.
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

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

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

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

Understanding the Electric Field of a Uniformly Charged Ring

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

Derivation of Equation of Trajectory Explained for Students

Understanding Electromagnetic Waves and Their Importance

