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Question Number 182613 by sciencestudent last updated on 11/Dec/22

	Answered by Acem last updated on 12/Dec/22

	$D i s . = v_{1} (t + 4) = v_{2} (t - 2)$ $\Rightarrow \frac{1}{3} (t + 4) = \frac{5}{12} (t - 2), t = 26 m i n_{\frac{5}{12} t h e s a m e}^{E x p \frac{1}{3} = \frac{v_{1}}{60} c o n v . t o m i n}$ $D i s = v_{1} T = 20 \times \frac{1}{2} = 10 k m$
	Answered by a.lgnaoui last updated on 12/Dec/22

	$s o i t x : d i s t s n c e t o s c h o o l$ $t_{1} = t e m p s m i s a l a v i t e s s e v_{1} = 20 k m / h$ $t_{1} = t + 4 / 60 (t : e n h e u r e s)$ $t_{2} = t e m p s m i s a l a v i t e s s e v_{2} = 25 k m / h$ $t_{2} = t - \frac{2}{60} v$ $t_{1} - \frac{1}{15} = t_{2} + \frac{1}{30} \Rightarrow t_{1} - t_{2} = \frac{1}{10}$ $t_{1} = t_{2} + \frac{1}{10}$ $x = v_{1} \times t_{1} = v_{2} \times t_{2} \Rightarrow \frac{t_{2}}{t_{1}} = \frac{4}{5} t_{1} = \frac{5 t_{2}}{4}$ $\frac{5 t_{2}}{4} = t_{2} + \frac{1}{10} t_{2} = \frac{2}{5} h$ $x = v_{2} \times t_{2} = 25 \times \frac{2}{5}$ $x = 10 k m$
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