Answer
64.8k+ views
Hint: The proceeding approach will be the relative motion concept. We going to assume boatman’s velocity with respect to water So that it will become easy to examine boatman’s motion in water
Before proceeding student must know what does relative approach mean, In most simple words, Relative motion is the understanding of the motion of an object with respect to some other moving or stationary object, for example a person is sitting in the moving bus is at zero velocity relative to the bus, is moving at the same velocity as the bus with respect to the ground.
Complete step by step solution:
According to the question given we have to find the minimum speed so that the boatman will reach at point B starting from A.
Let v be the velocity of a boatman with respect to water or velocity in still water, u is the velocity of water.
Resultant of v and u should be along AB.
Let the absolute velocity of boatman (velocity with respect to ground) be $v_b^ \to $ which is along AB
So, the resultant of u and $v_b^ \to $ should be along AB
Let us resolve component of v along x and y direction
Where ${v_x}$ and ${v_y}$ be the components along Horizontal and vertical directions respectively
So, coming to the calculation part
$\Rightarrow {v_x}$$ = u - v\sin \theta $
$\Rightarrow {v_y} = v\cos \theta $
Further $\tan \theta = \dfrac{{perpendicular}}{{base}} = \dfrac{{{v_y}}}{{{v_x}}}$
So, $\tan 45$
$
\Rightarrow \dfrac{{v\cos \theta }}{{u - v\sin \theta }} = 1\,\,\,\,\,\,\,\,(\tan 45 = 1) \\
\Rightarrow v\cos \theta = u - v\sin \theta \\
\therefore v = \dfrac{u}{{\sin \theta + \cos \theta }} \\
\Rightarrow v = \dfrac{u}{{\sqrt 2 \sin (\theta + 45)}} \\
$
So, v is minimum when the denominator is maximum, In denominator we have sin whose maximum value is 1.
$
\Rightarrow \theta + 45 = 90 \\
\Rightarrow \theta = {45^o} \\
$
And, $v = \dfrac{u}{{\sqrt 2 }}$
Speed relative to water is $\dfrac{u}{{\sqrt 2 }}$.
Direction is 45 degrees to the north west.
Note: The question above is for minimum speeds but for minimum distance we should move in such a direction that our motion will become along AC relative to water and for minimum time motion will be along AC direct.
Before proceeding student must know what does relative approach mean, In most simple words, Relative motion is the understanding of the motion of an object with respect to some other moving or stationary object, for example a person is sitting in the moving bus is at zero velocity relative to the bus, is moving at the same velocity as the bus with respect to the ground.
Complete step by step solution:
According to the question given we have to find the minimum speed so that the boatman will reach at point B starting from A.
Let v be the velocity of a boatman with respect to water or velocity in still water, u is the velocity of water.
Resultant of v and u should be along AB.
Let the absolute velocity of boatman (velocity with respect to ground) be $v_b^ \to $ which is along AB
So, the resultant of u and $v_b^ \to $ should be along AB
Let us resolve component of v along x and y direction
Where ${v_x}$ and ${v_y}$ be the components along Horizontal and vertical directions respectively
So, coming to the calculation part
$\Rightarrow {v_x}$$ = u - v\sin \theta $
$\Rightarrow {v_y} = v\cos \theta $
Further $\tan \theta = \dfrac{{perpendicular}}{{base}} = \dfrac{{{v_y}}}{{{v_x}}}$
So, $\tan 45$
$
\Rightarrow \dfrac{{v\cos \theta }}{{u - v\sin \theta }} = 1\,\,\,\,\,\,\,\,(\tan 45 = 1) \\
\Rightarrow v\cos \theta = u - v\sin \theta \\
\therefore v = \dfrac{u}{{\sin \theta + \cos \theta }} \\
\Rightarrow v = \dfrac{u}{{\sqrt 2 \sin (\theta + 45)}} \\
$
So, v is minimum when the denominator is maximum, In denominator we have sin whose maximum value is 1.
$
\Rightarrow \theta + 45 = 90 \\
\Rightarrow \theta = {45^o} \\
$
And, $v = \dfrac{u}{{\sqrt 2 }}$
Speed relative to water is $\dfrac{u}{{\sqrt 2 }}$.
Direction is 45 degrees to the north west.
Note: The question above is for minimum speeds but for minimum distance we should move in such a direction that our motion will become along AC relative to water and for minimum time motion will be along AC direct.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)