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A bus travels 6 km towards north at an angle of 45^(@)to the east and then travels 4 km towards north at an angle of 135^(@)to the east . How for is its final position , due east and due north ? How far is the point from the starting point ? What angle does the straight line joining its initial andfinal position makes with the east ?

Answer» <html><body><p></p>Solution :Net movement along x-direction (<a href="https://interviewquestions.tuteehub.com/tag/due-433472" style="font-weight:bold;" target="_blank" title="Click to know more about DUE">DUE</a> east)<br/>`= (6 - <a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>) cos 45^(@)`<br/>`= 2 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> (1)/(sqrt(2)) = sqrt(2) km`<br/>Net movement along y-direction (due <a href="https://interviewquestions.tuteehub.com/tag/north-1124498" style="font-weight:bold;" target="_blank" title="Click to know more about NORTH">NORTH</a>)<br/>`= ( 6 + 4) sin 45^(@)`<br/>`= 10 xx (1)/(sqrt(2)) = 5 sqrt(2) km`<br/>Netmovement from starting point<br/>= 6 + 4 = 10 km<br/>Angle which makes with the east direction <br/>`<a href="https://interviewquestions.tuteehub.com/tag/tan-1238781" style="font-weight:bold;" target="_blank" title="Click to know more about TAN">TAN</a> theta = ("y-component")/("x-component") = (5 sqrt(2))/(sqrt(2))`<br/>`:. theta = tan^(-1) (5) `(or)<br/>`theta = 79^(@) `(approx).<br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PRE_GRG_PHY_XI_V02_C02_E02_255_S01.png" width="80%"/></body></html>


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