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A car travles 6km towards north at an angle of `45^(@)` to the east and then travles distance of 4km towards north at an angle of `135^(@)` to east (figure). 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? |
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Answer» Net movement along x-direction, `S_(x)=(6-4) cos 45^(@) hat(i)=2xx1/sqrt(2)=sqrt(2)km` Net movement along y-direction, `S_(y)=(6+4) sin 45^(@) hat(j)=10xx1/sqrt(2)=5sqrt(2)km` Net movement from starting point, `|vec(s)|=sqrt(s_(a)^(2)+s_(y)^(2))=sqrt((sqrt(2))+(5sqrt(2))^(2))=sqrt(52) km` Angle which makes with the east direction, `tan theta=(Y-compon ent)/(X-compon ent)=(5sqrt(2))/sqrt(2) :. theta=tan^(-1)(5)` |
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