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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 $135^o$ to 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 ${\tan ^{ - 1}}\,\left( 5 \right)$
$10\, km$ and ${\tan ^{ - 1}}\,\left( {\sqrt 5 } \right)$
$\sqrt {52} \,km$ and ${\tan ^{ - 1}}\,\left( 5 \right)$
$\sqrt {52} \,km$ and ${\tan ^{ - 1}}\,\left( {\sqrt 5 } \right)$
Solution
$\begin{array}{l}
Net\,{\rm{ distance}}\,travellel\,along\,y – direction\\
{s_y} = 6\,\cos \,{45^ \circ }\hat j – 4\,\cos \,{45^ \circ }\hat j\\
= 10 \times \frac{1}{{\sqrt 2 }} = 5\sqrt 2 \,km\\
\therefore \,Net\,{\rm{distance}}\,travelled\,from\,the\,starting\\
{\rm{point}},
\end{array}$
$\begin{array}{l}
S = \sqrt {{s_x}^2 + {s_y}^2} = \sqrt {{{\left( {\sqrt 2 } \right)}^2} + {{\left( {5\sqrt 2 } \right)}^2}} \\
= \sqrt {2 + 25 \times 2} = \sqrt {52} \,km\\
Angle\,which\,the\,r{\rm{esultanl}}\,{\rm{makes}}\,{\rm{with}}\,{\rm{the}}\\
{\rm{east}}\,{\rm{direction}}\\
\tan \theta = \frac{y}{x} = \frac{{5\sqrt 2 }}{{\sqrt 2 }}\,\,\,\,\,\,\,\,\,or\,\theta \, = {\tan ^{ – 1}}\left( 5 \right)\,
\end{array}$
