Answer
Verified
385.8k+ views
Hint To find the angle between the direction of swimming of the man and that of the bank, first find the angle between the direction of swimming and the vertical. Find the component of the swimming of the man to the vertical to find the time taken to cross the river.
Formula used: In this solution we will be using the following formulae;
\[v = \dfrac{d}{t}\] where \[v\] is the velocity of a body in a particular direction, \[d\] is the displacement traversed by the body in a direction, and \[t\] is the time elapsed for the displacement to be covered.
\[hy{p^2} = op{p^2} + ad{j^2}\] where \[hyp\] signifies hypotenuse side of a right angled triangle, \[opp\] is opposite side, and \[adj\] is the adjacent side.
Complete Step-by-Step solution:
To find the angle between the targeted velocity of the swimming of the man (3km/h) and that of the velocity of the river, we must calculate the angle \[\theta \] first. This can be calculated as in
\[\tan \theta = \dfrac{2}{3}\]
\[ \Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{2}{3}} \right) = 41.7^\circ \]
Now, since, the river flow is parallel to the river bank, then the angle the man should try to swim
\[\alpha = 90 + \theta = 131.7^\circ \]
To find the time taken to cross the river, we must find the actual velocity of the man. We can use Pythagoras theorem, which says
\[hy{p^2} = op{p^2} + ad{j^2}\] where \[hyp\] signifies hypotenuse side of a right angled triangle, \[opp\] is opposite side, and \[adj\] is the adjacent side.
Hence,
\[{x^2} = {3^2} - {2^2} = 9 - 4 = 5\]
\[ \Rightarrow x = \sqrt 5 km/h\]
Now, from the equation
\[v = \dfrac{d}{t}\] where \[v\] is the velocity of a body in a particular direction, \[d\] is the displacement traversed by the body in a direction, and \[t\] is the time elapsed for the displacement to be covered, we have
\[\sqrt 5 = \dfrac{{0.5}}{t}\] (since 500 m is \[0.5km\])
Hence,
\[t = \dfrac{{0.5}}{{\sqrt 5 }} = 0.22hr\] or \[13.4\] minutes.
Note: Alternatively, one could calculate the real velocity of the man by finding the component of the 3km/h on the vertical axis, as in
\[v = 3\cos 41.7^\circ = 2.23\]
This is equivalent to \[\sqrt 5 \]
Formula used: In this solution we will be using the following formulae;
\[v = \dfrac{d}{t}\] where \[v\] is the velocity of a body in a particular direction, \[d\] is the displacement traversed by the body in a direction, and \[t\] is the time elapsed for the displacement to be covered.
\[hy{p^2} = op{p^2} + ad{j^2}\] where \[hyp\] signifies hypotenuse side of a right angled triangle, \[opp\] is opposite side, and \[adj\] is the adjacent side.
Complete Step-by-Step solution:
To find the angle between the targeted velocity of the swimming of the man (3km/h) and that of the velocity of the river, we must calculate the angle \[\theta \] first. This can be calculated as in
\[\tan \theta = \dfrac{2}{3}\]
\[ \Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{2}{3}} \right) = 41.7^\circ \]
Now, since, the river flow is parallel to the river bank, then the angle the man should try to swim
\[\alpha = 90 + \theta = 131.7^\circ \]
To find the time taken to cross the river, we must find the actual velocity of the man. We can use Pythagoras theorem, which says
\[hy{p^2} = op{p^2} + ad{j^2}\] where \[hyp\] signifies hypotenuse side of a right angled triangle, \[opp\] is opposite side, and \[adj\] is the adjacent side.
Hence,
\[{x^2} = {3^2} - {2^2} = 9 - 4 = 5\]
\[ \Rightarrow x = \sqrt 5 km/h\]
Now, from the equation
\[v = \dfrac{d}{t}\] where \[v\] is the velocity of a body in a particular direction, \[d\] is the displacement traversed by the body in a direction, and \[t\] is the time elapsed for the displacement to be covered, we have
\[\sqrt 5 = \dfrac{{0.5}}{t}\] (since 500 m is \[0.5km\])
Hence,
\[t = \dfrac{{0.5}}{{\sqrt 5 }} = 0.22hr\] or \[13.4\] minutes.
Note: Alternatively, one could calculate the real velocity of the man by finding the component of the 3km/h on the vertical axis, as in
\[v = 3\cos 41.7^\circ = 2.23\]
This is equivalent to \[\sqrt 5 \]
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Why Are Noble Gases NonReactive class 11 chemistry CBSE
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
At which age domestication of animals started A Neolithic class 11 social science CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Write a letter to the principal requesting him to grant class 10 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE