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Question Number 181232 by mr W last updated on 23/Nov/22

Commented by mr W last updated on 23/Nov/22

$a t r a i n d r i v e s f r o m A i n d i r e c t i o n t o$ $D w i t h s p e e d 160 k m / h . a t t h e s a m e$ $t i m e a c a r d r i v e s f r o m B i n d i r e c t i o n$ $t o E w i t h s p e e d 100 k m / h .$ $a f t e r w h a t t i m e i s t h e d i s t a n c e$ $b e t w e e n t h e m m i n i n u m a n d w h a t i s$ $t h i s m i n i m u m d i s t a n c e ?$ $(w e t r e a t b o t h t r a i n a n d c a r a s p o i n t s)$
	Answered by mahdipoor last updated on 23/Nov/22

	$b e f o r h = \frac{1}{10} (a t h = \frac{1}{10} t r a i n B i n C),$ $d (= d i s t a n c e a m o n g t r a i n s) w a s i n c r e a s$ $a f t e r h = \frac{1}{8} (t r a i n A i n C), d w i l l i n c r e a s$ $a m o n g 1 / 8 ⩾ h ⩾ 1 / 10 \Rightarrow$ $d^{2} = {(100 h - 10)}^{2} + {(20 - 160 h)}^{2}$ $- 2 \times (100 h - 10) (20 - 160 h) c o s 60$ $d (d^{2}) / d h = 2 (100 h - 10) 100 - 2 (20 - 160 h) 160$ $- 100 (20 - 160 h) + 160 (100 h - 10) =$ $360 (100 h - 10) - 420 (20 - 160 h) = 0$ $\Rightarrow 1032 h - 120 = 0 \Rightarrow h = \frac{5}{43} \approx . 12$ $\Rightarrow 0.125 > 0.12 > 0.1 \Rightarrow$ $m i n (d^{2}) = d^{2} (h = \frac{5}{43}) = \frac{100}{43} {k m}^{2}$ $\Rightarrow m i n d = \frac{10}{\sqrt{43}} \approx 1.525 k m$
	Commented by mr W last updated on 23/Nov/22

	$t h a n k s s i r!$
	Commented by mahdipoor last updated on 23/Nov/22

	$t n x, i e d i t e d$
	Answered by mr W last updated on 23/Nov/22

	Commented by mr W last updated on 23/Nov/22

	$\underset{―}{U s i n g m e t h o d o f r e l a t i v e m o t i o n}$ $a s s u m e t h e c a r {d o e s n}^{'} t m o v e . t h e n$ $t h e t r a i n m o v e s w i t h t h e r e l a t i v e$ $v e l o c i t y v i n d i r e c t i o n F, s e e p i c t u r e .$ $t h e s h o r t e s t d i s t a n c e b e t w e e n c a r$ $a n d t r a i n i s t h e d i s t a n c e f r o m B t o$ $t h e l i n e A F .$ $v = \sqrt{160^{2} + 100^{2} + 160 \times 100} = 20 \sqrt{129} k m / h$ $\sin α = \frac{100}{20 \sqrt{129}} \times \sin 120 = \frac{5}{2 \sqrt{43}}$ $β = 90 ° - α$ $\sin β = \cos α = \frac{7 \sqrt{3}}{2 \sqrt{43}}$ $\frac{B Q}{\sin 60 °} = \frac{C Q}{\sin (60 ° + β)} = \frac{C B}{\sin β}$ $B Q = \frac{10 \times 2 \sqrt{43}}{7 \sqrt{3}} \times \frac{\sqrt{3}}{2} = \frac{10 \sqrt{43}}{7}$ $C Q = \frac{10 \times 2 \sqrt{43}}{7 \sqrt{3}} \times (\frac{\sqrt{3}}{2} \times \frac{5}{2 \sqrt{43}} + \frac{1}{2} \times \frac{7 \sqrt{3}}{2 \sqrt{43}})$ $= \frac{60}{7}$ $P B = (20 + \frac{60}{7}) \sin α - \frac{10 \sqrt{43}}{7}$ $= \frac{10}{\sqrt{43}} \approx 1.525 k m$ $t = \frac{A P}{v} = \frac{1}{20 \sqrt{129}} (20 + \frac{60}{7}) \cos α$ $= \frac{5}{43} \approx 0.116 h = 6 m i n 58.6 s e c$
	Commented by Acem last updated on 23/Nov/22

	Commented by Acem last updated on 23/Nov/22

	$T h e q u s e t i o n i s v e r y g o o d, t h a n k y o u f o r i t S i r .$ $W e l l, p l e a s e c h e c k t h e d i f f e r e n c e b e t w e e n$ $2.32 a n d 1.538 k m$ $2 n d : a m c o n f u s e d t h a t w h y i g o t$ $7 m i n : 2330 m e t e r$ $b u t t i m e 6 m i n 58.6 s e c : 2320 m e t e r$ ${t h e r e}^{'} s a d i f f e r e n c e a b o u t 10 m e t e r$ $N o t e : i w i l l w r i t e m i n e w h e n f i n i s h m y w o r k$
	Commented by mr W last updated on 23/Nov/22

	$c o r r e c t a n s w e r :$ $a f t e r \frac{5}{43} h, i . e . 6 m i n 58.6 s e c t h e$ $s h o r t e s t d i s t a n c e \frac{10}{\sqrt{43}} \approx 1.525 k m .$
	Answered by mr W last updated on 23/Nov/22

	Commented by mr W last updated on 23/Nov/22

	$\underset{―}{A n o t h e r g e o m e t r i c a l w a y}$ $s a y w h e n t r a i n i s a t p o i n t A_{1} a n d$ $c a r i s a t p o i n t B_{1} t h e i r d i s t a n c e i s$ $m i n i m u m, a t t h i s m o m e n t t h e r a t e$ $i n w h i c h t h e i r d i s t a n c e c h a n g e s i s$ $z e r o :$ $160 \cos θ_{1} - 100 \cos θ_{2} = 0$ $\Rightarrow \cos θ_{1} = \frac{5}{8} \cos θ_{2}$ $θ_{1} + θ_{2} = 180 ° - 60 ° = 120 °$ $\cos θ_{1} = \cos (120 ° - θ_{2}) = - \frac{1}{2} \cos θ_{2} + \frac{\sqrt{3}}{2} \sin θ_{2}$ $\frac{5}{8} \cos θ_{2} = - \frac{1}{2} \cos θ_{2} + \frac{\sqrt{3}}{2} \sin θ_{2}$ $\Rightarrow \tan θ_{2} = \frac{9}{4 \sqrt{3}}$ $\Rightarrow \cos θ_{2} = \frac{4}{\sqrt{43}} \Rightarrow \cos θ_{1} = \frac{5}{2 \sqrt{43}}$ $\Rightarrow \sin θ_{1} = \frac{7 \sqrt{3}}{2 \sqrt{43}}, \sin θ_{2} = \frac{3 \sqrt{3}}{\sqrt{43}}$ $A_{1} C = 20 - 160 t$ $B_{1} C = 100 t - 10$ $\frac{A_{1} C}{\sin θ_{2}} = \frac{B_{1} C}{\sin θ_{1}} = \frac{A_{1} B_{1}}{\sin 60 °}$ $\frac{\sqrt{43} (20 - 160 t)}{3 \sqrt{3}} = \frac{2 \sqrt{43} (100 t - 10)}{7 \sqrt{3}} = \frac{A_{1} B_{1}}{\sin 60 °}$ $\frac{\sqrt{43} (20 - 160 t)}{3 \sqrt{3}} = \frac{2 \sqrt{43} (100 t - 10)}{7 \sqrt{3}}$ $\Rightarrow t = \frac{5}{43} ✓$ $A_{1} B_{1} = \frac{\sqrt{43} (20 - 160 \times \frac{5}{43})}{3 \sqrt{3}} \times \frac{\sqrt{3}}{2} = \frac{10}{\sqrt{43}} ✓$
	Commented by Acem last updated on 23/Nov/22

	Commented by mr W last updated on 23/Nov/22

	$y o u m i s c a l c u l a t e d L .$
	Commented by Acem last updated on 23/Nov/22

	$\sqrt{h o w a m i ?}$ $W e b o t h h a v e t h e s a m e r e s u l t s! t h a n k y o u$ $a f t e r a l i t t l e w i l l w r i t e t h e m e t h o d$
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