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
263.7k+ 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 2025-26 Experimental Skills Mock Test – Free Practice

JEE Main 2025-26 Electronic Devices Mock Test: Free Practice Online

JEE Main 2025-26 Mock Tests: Free Practice Papers & Solutions

JEE Main 2025-26: Magnetic Effects of Current & Magnetism Mock Test

JEE Main 2025-26 Atoms and Nuclei Mock Test – Free Practice Online

JEE Main Mock Test 2025-26: Experimental Skills Chapter Online Practice

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Kinematics Mock Test for JEE Main 2025-26: Comprehensive Practice

Kinematics Mock Test for JEE Main 2025-26: Practice & Ace the Exam

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

JEE Advanced Marks vs Rank 2025 - Predict Your IIT Rank Based on Score

Derivation of Equation of Trajectory Explained for Students

