# A motor boat covers the distance between two spots on the river banks in ${t_1} = 8h$ and ${t_2} = 12h$ downstream and upstream respectively. The time required for the boat to cover this distance in still water will be?

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**Hint:**In order to answer this problem let us first get some idea about Relative Velocity. The relative velocity is the speed at which body $A$ appears to an observer on body B, and vice versa. The vector difference between the velocities of two bodies is the relative velocity in mathematics. \[{{\text{V}}_{AB}}{\text{ }} = {\text{ }}{{\text{v}}_A} - {v_B}\] is the equation.

**Complete step by step answer:**

Let us get some ideas about rivers and boats to solve this problem. The wind's effect on the plane is comparable to the river current's effect on the motorboat. A motorboat would not meet the shore directly across from its starting point if it were to head straight across a river (that is, if the boat were to point its bow straight towards the other side). The boat's motion is influenced by the river current, which takes it downstream.

To solve river boat problems, we need to understand two concepts:

-A boat's speed in relation to the sea is the same as a boat's speed in still water.

-The velocity of a boat relative to the water (${v_{b/w}}$) is equal to the difference in velocity of the boat relative to the ground (${v_b}$) and velocity of water relative to the ground (${v_w}$) i.e.

${v_{b/w}} = {v_b} - {v_w}$

Let apply this concept to our problem: let the time needed in still water be $t$ $hr$ and the boat's speed be \[v\] and ${v^1}$.For still water

$d = vt$..................$[equation - 1]$

where, $d$=distance, $v$=velocity and $t$=time.

For downstream

$d = (v + {v^1}) \times 8$...................$[equation - 2]$

For upstream

$d = (v - {v^1}) \times 12$....................$[equation - 3]$

Solving $2$ & $3$ we get

$3 \times (v - {v^1}) = 2 \times (v + {v^1})$

$\Rightarrow v = 5{v^1}$.....................$[equation - 4]$

Putting value of $v$ in $equ - 2$ and also value of $d$ from $equ - 1$

$vt = (v + \dfrac{v}{5}) \times 8$

$\Rightarrow t = \dfrac{{6 \times 8}}{5} \\

\Rightarrow t= \dfrac{{48}}{5} \\

\therefore t= 9.6\,hr$

**Note:**To solve problems based on boat and stream above two concepts are necessary to keep in mind. If you are going to remember the concept then it will be very easy for you to solve the problem.

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