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A car travels $6 \,km$ towards north at an angle of $45^o $ to the east and then travels distance of $4 \,km$ towards north at an angle of $135^o $ to the east. How far is the point from the starting point. What angle does the straight line joining its initial and final position makes with the east
$\sqrt {50} \,km$ and ${\tan ^{ - 1}}(5)$
$10 \,km$ and ${\tan ^{ - 1}}(\sqrt 5 )$
$\sqrt {52} \,km$ and ${\tan ^{ - 1}}(5)$
$\sqrt {52} \,km$ and ${\tan ^{ - 1}}(\sqrt 5 )$
Solution
(c) Net movement along $x-$direction $Sx = (6 -4) cos 45^o $ $\hat i$
$ = 2 \times \frac{1}{{\sqrt 2 }}\, = \,\sqrt 2 \,km$
Net movement along y-direction $Sy = (6 + 4) sin 45^o $$\hat j$ $ = 10 \times \frac{1}{{\sqrt 2 }} = 5\sqrt 2 \,km$
Net movement from starting point
$|\overrightarrow S |\, = \sqrt {{S_x}^2 + {S_y}^2} $$ = \sqrt {{{\left( {\sqrt 2 } \right)}^2} + {{\left( {5\sqrt 2 } \right)}^2}} $=$\sqrt {52} \,\,km$
Angle which makes with the east direction
$\tan \theta = \frac{{{\rm{Y}} – {\rm{component}}}}{{{\rm{X}} – \,{\rm{component}}}}$ $ = \frac{{5\sqrt 2 }}{{\sqrt 2 }}$ $\therefore \theta = {\tan ^{ – 1}}(5)$
