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Theorems for Continuity

Car A is mo...

Question

Car $A$ is moving due east with speed $v_{1}$ . At a certain time it is $l_{1}$ away from the crossing.Car $B$ at that instant is $l_{2}$ away from the crossing and moving due north with a speed $v_{2}$ . Find the shortest distance between them.

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Solution

$D i s t a n c e o f c a r A f r o m t h e c r o s s i n g a t a n y t i m e t = l_{1} - v_{1} t D i s t a n c e o f c a r B f r o m t h e c r o s s i n g a t a n y t i m e t = l_{2} - v_{2} t D i s t a n c e b e t w e e n t h e s e t w o c a r s = \sqrt{{(l_{1} - v_{1} t)}^{2} + {(l_{2} - v_{2} t)}^{2}} = s \frac{d s}{d t} = \frac{- 2 (l_{1} - v_{1} t) v_{1} - 2 (l_{2} - v_{2} t) v_{2}}{2 \sqrt{{(l_{1} - v_{1} t)}^{2} + {(l_{2} - v_{2} t)}^{2}}} = 0 \Rightarrow t = \frac{l_{1} v_{1} + l_{2} v_{2}}{v_{1}^{2} + v_{2}^{2}} m i n d i s t a n c e b e t w e e n t h e m i s s_{m i n} = v_{1} v_{2} \sqrt{\frac{l_{1}^{2} + l_{2}^{2}}{v_{1}^{2} + v_{2}^{2}}}$

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