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In-a-race-on-a-track-of-a-certain-length-A-beats-B-by-20-m-When-A-was-10-m-ahead-of-the-mid-point-of-the-track-B-was-2-m-behind-it-Find-the-length-of-the-track-in-m-




Question Number 24428 by dhandhaliyachandra@gmail.Com last updated on 17/Nov/17
In a race on a track  of a certain length,  A  beats B by 20 m. When A was 10 m  ahead of the mid−point of the track B  was 2 m behind it. Find the length of the  track (in m).
$$\mathrm{In}\:\mathrm{a}\:\mathrm{race}\:\mathrm{on}\:\mathrm{a}\:\mathrm{track}\:\:\mathrm{of}\:\mathrm{a}\:\mathrm{certain}\:\mathrm{length}, \\ $$$$\mathrm{A}\:\:\mathrm{beats}\:\mathrm{B}\:\mathrm{by}\:\mathrm{20}\:\mathrm{m}.\:\mathrm{When}\:\mathrm{A}\:\mathrm{was}\:\mathrm{10}\:\mathrm{m} \\ $$$$\mathrm{ahead}\:\mathrm{of}\:\mathrm{the}\:\mathrm{mid}−\mathrm{point}\:\mathrm{of}\:\mathrm{the}\:\mathrm{track}\:\mathrm{B} \\ $$$$\mathrm{was}\:\mathrm{2}\:\mathrm{m}\:\mathrm{behind}\:\mathrm{it}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{track}\:\left(\mathrm{in}\:\mathrm{m}\right). \\ $$
Answered by ajfour last updated on 17/Nov/17
(v_A −v_B )T=20   ;  v_A T = l  (((l/2)+10)/v_A ) = (((l/2)−2)/v_B )  = (t say)  ⇒ from first two eqns.  1−(v_B /v_A ) = ((20)/l)    or     (v_B /v_A ) = 1−((20)/l)  using this in third  eqn.  (((l/2)+10)/1) =(((l/2)−2)/((1−((20)/l))))  ⇒  (l+20)(l−20)=l(l−4)      l^2 −400 = l^2 −4l  ⇒     l = 100 m .
$$\left({v}_{{A}} −{v}_{{B}} \right){T}=\mathrm{20}\:\:\:;\:\:{v}_{{A}} {T}\:=\:{l} \\ $$$$\frac{\left({l}/\mathrm{2}\right)+\mathrm{10}}{{v}_{{A}} }\:=\:\frac{\left({l}/\mathrm{2}\right)−\mathrm{2}}{{v}_{{B}} }\:\:=\:\left({t}\:{say}\right) \\ $$$$\Rightarrow\:{from}\:{first}\:{two}\:{eqns}. \\ $$$$\mathrm{1}−\frac{{v}_{{B}} }{{v}_{{A}} }\:=\:\frac{\mathrm{20}}{{l}}\:\:\:\:{or}\:\:\:\:\:\frac{{v}_{{B}} }{{v}_{{A}} }\:=\:\mathrm{1}−\frac{\mathrm{20}}{{l}} \\ $$$${using}\:{this}\:{in}\:{third}\:\:{eqn}. \\ $$$$\frac{\left({l}/\mathrm{2}\right)+\mathrm{10}}{\mathrm{1}}\:=\frac{\left({l}/\mathrm{2}\right)−\mathrm{2}}{\left(\mathrm{1}−\frac{\mathrm{20}}{{l}}\right)} \\ $$$$\Rightarrow\:\:\left({l}+\mathrm{20}\right)\left({l}−\mathrm{20}\right)={l}\left({l}−\mathrm{4}\right) \\ $$$$\:\:\:\:{l}^{\mathrm{2}} −\mathrm{400}\:=\:{l}^{\mathrm{2}} −\mathrm{4}{l} \\ $$$$\Rightarrow\:\:\:\:\:{l}\:=\:\mathrm{100}\:{m}\:. \\ $$$$ \\ $$

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