On a two lane road,car A is travelling with a speed of 36km/h. Two cars B and C approach A in opposite directions with a speed of 54km/h each.At a certain instant, when the distance AB is equal to AC,both being 1km,B decides to overtake A before C does.What minimum acceleration of carB is required to avoid an accident?


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Question Number 29359 by NECx last updated on 08/Feb/18

$O n a t w o l a n e r o a d, c a r A i s$ $t r a v e l l i n g w i t h a s p e e d o f 36 k m / h .$ $T w o c a r s B a n d C a p p r o a c h A i n$ $o p p o s i t e d i r e c t i o n s w i t h a s p e e d o f$ $54 k m / h e a c h . A t a c e r t a i n i n s t a n t,$ $w h e n t h e d i s t a n c e A B i s e q u a l$ $t o A C, b o t h b e i n g 1 k m, B d e c i d e s$ $t o o v e r t a k e A b e f o r e C d o e s . W h a t$ $m i n i m u m a c c e l e r a t i o n o f c a r B$ $i s r e q u i r e d t o a v o i d a n a c c i d e n t ?$
	Answered by 33 last updated on 08/Feb/18

	$v_{r e l C} = 15 - (- 10) = 25 m / s$ $v_{r e l B} = 15 - 10 = 5 m / s$ $B m u s t r e a c h A j u s t b e f o r e C$ $r e a c h e s t o a v o i d a c c i d e n t .$ $l e t t_{C} ⋍ t_{B}$ $t_{C} = \frac{1000}{25} = 40 s (S_{r e l} / v_{r e l})$ $u s i n g s e c o n d e q u a t i o n o f m o t i o n$ $1000 = 5 (40) + \frac{1}{2} (a) (40^{2})$ $\Rightarrow a = 1 m / s^{2} i f t h e y r e a c h A$ $s i m u l t a n e o u s l y .$ $\Rightarrow a ⪈ 1 m / s^{2}$
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