Answer

Verified

408k+ 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 \[\sqrt 5 \]

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

The polyarch xylem is found in case of a Monocot leaf class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Casparian strips are present in of the root A Epiblema class 12 biology CBSE

How do you graph the function fx 4x class 9 maths CBSE