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Question Number 216621 by Tawa11 last updated on 12/Feb/25

	Answered by A5T last updated on 12/Feb/25

	Commented by A5T last updated on 12/Feb/25

	$[{OA}_{1} DT] = [ABCT] when both cars meet at time t$ $\Rightarrow 20 t = \frac{30 (t - 20 + t - 10)}{2} \Rightarrow 4 t = 3 (2 t - 30)$ $\Rightarrow t = 45 s \Rightarrow s = 20 \times 45 = 900 m$ $\Rightarrow Sports car overtakes first car 45 s and 900 m$ $after car overtakes stationary sports car .$
	Commented by Tawa11 last updated on 12/Feb/25

	$Thanks sir .$ $I appreciate .$
	Answered by mr W last updated on 12/Feb/25

	Commented by mr W last updated on 12/Feb/25

	$d i s t a n c e i s t h e a r e a u n d e r t h e$ $c o r r e s p o n d i n g v - t c u r v e .$ $d = 20 x = \frac{30 \times 10}{2} + 30 (x - 20)$ $\Rightarrow x = 45 s$ $\Rightarrow d = 20 \times 45 = 900 m$
	Commented by Tawa11 last updated on 12/Feb/25

	$Thanks sir .$ $I appreciate .$
	Commented by Tawa11 last updated on 14/Feb/25

	$Sirs, can this be solved without graph ?$
	Commented by mr W last updated on 15/Feb/25

	$t h e s o l u t i o n a b o v e i s w i t h o u t$ $g r a p h! o r y o u {d o n}^{'} t u n d e r s t a n d i t$ $w i t h o u t t h e g r a p h ?$
	Commented by Tawa11 last updated on 15/Feb/25

	$I mean, can we solve analytically$ $without solving from graph ?$ $Can we solve the distance and time$ $without drawing graph (v - t graph) ?$
	Commented by mr W last updated on 15/Feb/25

	$t h e y a r e e x a c t l y g i v e n a b o v e .$ $d_{1} = 20 x$ $d_{2} = \frac{30 \times 10}{2} + 30 (x - 20)$ $w i t h x = t i m e$
	Commented by Tawa11 last updated on 15/Feb/25

	$Ohh, I grab sir .$ $Thanks sir .$
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