A river is flowing due east with a speed 3m/s. A swimmer can swim in still water at a speed of 4 m/s .

(a) If swimmer starts swimming due north, what will be his resultant velocity (magnitude and direction)?

(b) If he wants to start from point A on south bank and reach opposite point B on north bank,

(a) Which direction should he swim?

(b) What will be his resultant speed?

(c) From two different cases as mentioned in (a) and (b) above, in which case will he reach opposite bank in shorter time?

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation – speed of river V_r= 3m/s

Speed of swimmer V_s= 4m/s

(a) when swimmer starts swimming due north then its resultant velocity

V= $\sqrt[]{v_r^2+v_s^2}=\sqrt[]{3^2+4^2}=5m/s$

tan so ’N

(b) to reach at point B resultant velocity will be

V= $\sqrt[]{v_s^2-v_r^2}=\sqrt[]{4^2-3^2}=\sqrt[]{7}m/s$

tan $\theta =\frac{v_r}{v}=\frac{3}{\sqrt[]{7}}$

(c) time taken by swimmer t =d/v= d/4s

in case b time taken by swimmer to cross the river

t₁=d/v=d/ $\sqrt[]{7}$

so t1 

<p>This is a Long Answer Type Question as classified in NCERT Exemplar</p><p><strong>Explanation</strong> – speed of river V_r= 3m/s</p><p>Speed of swimmer V_s= 4m/s</p><p> (a) when swimmer starts swimming due north then its resultant velocity</p><p>V=<span contenteditable="false"> <math> <mroot> <mrow> <mrow> <msubsup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> <mo>+</mo> <msubsup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>s</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> </mrow> </mrow> <mrow></mrow> </mroot> <mo>=</mo> <mroot> <mrow> <mrow> <msup> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow></mrow> </mroot> <mo>=</mo> <mn>5</mn> <mi>m</mi> <mo>/</mo> <mi>s</mi> </math> </span></p><p>tan&nbsp;so ’N</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745411031php13seV8_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745411031php13seV8.jpeg" alt="" width="175" height="126"></picture></div><div><p> (b) to reach at point B resultant velocity will be</p><p>V=<span contenteditable="false"> <math> <mroot> <mrow> <mrow> <msubsup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>s</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> <mo>-</mo> <msubsup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> </mrow> </mrow> <mrow></mrow> </mroot> <mo>=</mo> <mroot> <mrow> <mrow> <msup> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>-</mo> <msup> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow></mrow> </mroot> <mo>=</mo> <mroot> <mrow> <mrow> <mn>7</mn> </mrow> </mrow> <mrow></mrow> </mroot> <mi>m</mi> <mo>/</mo> <mi>s</mi> </math> </span></p><p>tan<span contenteditable="false"> <math> <mi>θ</mi> <mo>=</mo> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> <mrow> <mrow> <mroot> <mrow> <mrow> <mn>7</mn> </mrow> </mrow> <mrow></mrow> </mroot> </mrow> </mrow> </mfrac> </math> </span></p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745411100php0SdY3f_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745411031php13seV8.jpeg"7 alt="" width="157" height="116"></picture></div><div><p> (c) time taken by swimmer t =d/v= d/4s</p><p>in case b time taken by swimmer to cross the river</p><p>t₁=d/v=d/<span contenteditable="false"> <math> <mroot> <mrow> <mrow> <mn>7</mn> </mrow> </mrow> <mrow></mrow> </mroot> <mi> </mi> </math> </span></p><p>so t&lt;t_{1&nbsp;&lt;/t}</p></div></div></div></div>

0 Upvotes 0 Upvotes 0 Downvotes

• • Report Abuse
• Comment