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In a harbour ,wind is blowing at the speed of 27 km //hand the flag on the mast of a boat anchored in the harbour flutters along the N-E direction . If the boat starts moving at a speed of 51 km//h to the north , what is the direction of the flag on the mast of the boat ?

Answer» <html><body><p></p>Solution :As the flag <a href="https://interviewquestions.tuteehub.com/tag/flutters-2085792" style="font-weight:bold;" target="_blank" title="Click to know more about FLUTTERS">FLUTTERS</a> in the N-E direction , the wind must be flowing in the N-E direction. When the boat start moving , the flag will flutter in the direction ofrelative velocity of wind w.r.t. boat . <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/MOD_UNT_PHY_XI_P1_C04_E04_014_S01.png" width="80%"/> <br/>`|vecv_(w)|=<a href="https://interviewquestions.tuteehub.com/tag/72-334274" style="font-weight:bold;" target="_blank" title="Click to know more about 72">72</a> km //h` <br/> `|vecv_(b)|=<a href="https://interviewquestions.tuteehub.com/tag/51-324358" style="font-weight:bold;" target="_blank" title="Click to know more about 51">51</a> km//h` <br/> If `beta` is the angle between `vec v_(w) and -vecv_(b)` <br/> `tan beta =(51 sin135 ^(@))/(72+51 cos <a href="https://interviewquestions.tuteehub.com/tag/130-272294" style="font-weight:bold;" target="_blank" title="Click to know more about 130">130</a>^(@))` <br/> `=((51xx1 sqrt2))/(72-51xx1sqrt2)` <br/> `=131.1//51.1=0.9965` <br/> `therefore beta=tan^(-1) (1.0038)=45.1^(@)`(with S-E) <br/> `therefore` Angle with respect to direction of east<br/>`=45.1-45^(@)=0.1` (with east)<br/> Hence , the direction of the flag is almost towards east .</body></html>


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