
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
169.2k+ 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.
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