એક કાર ઉત્તર તરફ પૂર્વ દિશા સાથે 45^o ના કોણે | vedclass

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Question

$\begin{array}{l}
Net\,{m{ distance}}\,travellel\,along\,y - direction\
{s_y} = 6\,\cos \,{45^ \circ }\hat j - 4\,\cos \,{45^ \circ }\hat j\
&n

Vedclass · Accepted Answer

$\sqrt {52} \,km$ અને ${	an ^{ - 1}}\,\left( 5 ight)$

Vedclass · Answer

$\begin{array}{l}
Net\,{m{ distance}}\,travellel\,along\,y - direction\
{s_y} = 6\,\cos \,{45^ \circ }\hat j - 4\,\cos \,{45^ \circ }\hat j\
 = 10 	imes \frac{1}{{\sqrt 2 }} = 5\sqrt 2 \,km\
	herefore \,Net\,{m{distance}}\,travelled\,from\,the\,starting\
{m{point}},
\end{array}$

$\begin{array}{l}
S = \sqrt {{s_x}^2 + {s_y}^2}  = \sqrt {{{\left( {\sqrt 2 } ight)}^2} + {{\left( {5\sqrt 2 } ight)}^2}} \
 = \sqrt {2 + 25 	imes 2}  = \sqrt {52} \,km\
Angle\,which\,the\,r{m{esultanl}}\,{m{makes}}\,{m{with}}\,{m{the}}\
{m{east}}\,{m{direction}}\
	an 	heta  = \frac{y}{x} = \frac{{5\sqrt 2 }}{{\sqrt 2 }}\,\,\,\,\,\,\,\,\,or\,	heta \, = {	an ^{ - 1}}\left( 5 ight)\,
\end{array}$

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