A motor boat going downstream overcame a raft at a point $A$. $60\min $ later it turns back & after some time it passes a raft again at a distance $6km$ from point $A$.Find the flow of velocity assuming the speed of the boat is constant.
Answer
252.9k+ views
Hint: This question can easily be solved if we know the application of relative velocities. First of all we need to find the relative velocities of the motor boat and the raft. After that we need to find the distance covered by the system and then finally solve the equations obtained to get the required solution for the given question.
Complete step by step solution:
First of all let us assume the velocity of the motorboat to be${v_b}$and the velocity of the flow of water to be${v_w}$.
Now, let us assume the total distance travelled by the boat downstream to be$x$
Therefore, using speed distance relation we can write,
$\Rightarrow x = ({v_b} + {v_w})t$…………….. (i)
Now, when the boat turns back, let us assume the time taken to be${t_1}$
The total distance covered is$6km$.
Now, we get another equation as,
$\Rightarrow x - 6 = ({v_b} - {v_w}){t_1}$……………. (ii)
Also, form the question, we have $t = $$60\min = 1hr$
$\Rightarrow 6 = {v_w}(t + {t_1})$……………… (iii)
From equation (i), (ii) and (iii), we get,
$\Rightarrow ({v_b} + {v_w})t - {v_w}(t + {t_1}) = ({v_b} - {v_w}){t_1}$
$ \Rightarrow {v_b}t + {v_w}t - {v_w}t - {v_w}{t_1} = {v_b}{t_1} - {v_{_w}}{t_1}$
$ \Rightarrow {v_b}t = {v_b}{t_1}$
$\therefore t = {t_1}$
Now, from equation (iii), we have,
$\Rightarrow {v_w} = \dfrac{6}{{t + {t_1}}} = \dfrac{6}{{2t}}$
Now, we need to put the required values in the above equation. So, after putting the values, we get,
$\Rightarrow {v_w} = \dfrac{6}{{2 \times 1}} = 3km{h^{ - 1}}$
Therefore, the required velocity of the flow of water is $3km{h^{ - 1}}$.
Note: When a boat moves along the direction of the stream in that case we say that the movement of the boat is downstream. : When a boat moves along the opposite direction of the stream in that case we say that the movement of the boat is upstream.
Complete step by step solution:
First of all let us assume the velocity of the motorboat to be${v_b}$and the velocity of the flow of water to be${v_w}$.
Now, let us assume the total distance travelled by the boat downstream to be$x$
Therefore, using speed distance relation we can write,
$\Rightarrow x = ({v_b} + {v_w})t$…………….. (i)
Now, when the boat turns back, let us assume the time taken to be${t_1}$
The total distance covered is$6km$.
Now, we get another equation as,
$\Rightarrow x - 6 = ({v_b} - {v_w}){t_1}$……………. (ii)
Also, form the question, we have $t = $$60\min = 1hr$
$\Rightarrow 6 = {v_w}(t + {t_1})$……………… (iii)
From equation (i), (ii) and (iii), we get,
$\Rightarrow ({v_b} + {v_w})t - {v_w}(t + {t_1}) = ({v_b} - {v_w}){t_1}$
$ \Rightarrow {v_b}t + {v_w}t - {v_w}t - {v_w}{t_1} = {v_b}{t_1} - {v_{_w}}{t_1}$
$ \Rightarrow {v_b}t = {v_b}{t_1}$
$\therefore t = {t_1}$
Now, from equation (iii), we have,
$\Rightarrow {v_w} = \dfrac{6}{{t + {t_1}}} = \dfrac{6}{{2t}}$
Now, we need to put the required values in the above equation. So, after putting the values, we get,
$\Rightarrow {v_w} = \dfrac{6}{{2 \times 1}} = 3km{h^{ - 1}}$
Therefore, the required velocity of the flow of water is $3km{h^{ - 1}}$.
Note: When a boat moves along the direction of the stream in that case we say that the movement of the boat is downstream. : When a boat moves along the opposite direction of the stream in that case we say that the movement of the boat is upstream.
Recently Updated Pages
States of Matter Chapter For JEE Main Chemistry

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

Circuit Switching vs Packet Switching: Key Differences Explained

Mass vs Weight: Key Differences Explained for Students

[Awaiting the three content sources: Ask AI Response, Competitor 1 Content, and Competitor 2 Content. Please provide those to continue with the analysis and optimization.]

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

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

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

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

Other Pages
JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

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

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2025-26

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2025-26

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

CBSE Notes Class 11 Physics Chapter 4 - Laws of Motion - 2025-26

