Lesson One very twisted Travel and Distance problem
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<H2>One very twisted Travel and Distance problem</H2> <H3>Problem 1</H3>Alice and Bob started to walk towards each other's home and then back to theirs, with steady speeds. Alice passed by a bus station at 20 m away from her home, while at the same time Bob was passing by an abandoned old car. Afterwards, they met at 55 m away from Bob's home and then they met again at 85 m from Alice's home. What is the distance between the bus station and the abandoned car? <B>Solution</B> <pre> Let S be the distance between their homes, in meters. Let {{{V[1]}}} be Alice's rate, and {{{V[2]}}} be Bob's rate. Let {{{t[1]}}} be the time after their start to their first meeting. Let {{{t[2]}}} be the time after their start to their second meeting. Then you have {{{V[1]*t[1]}}} = S - 55; {{{V[2]*t[1]}}} = 55; (1) {{{V[1]*t[2]}}} = 2S - 85; {{{V[2]*t[2]}}} = S + 85. (2) From (1) you have {{{V[1]/V[2]}}} = {{{(S-55)/55}}}; (3) From (2) you have {{{V[1]/V[2]}}} = {{{(2S-85)/(S+85)}}}, (4) which gives you an equation for S {{{(S-55)/55}}} = {{{(2S-85)/(S+85)}}}. Simplify and solve for S: (S-55)*(S+85) = (2S-85)*55 S^2 -55*S + 85*S - 55*85 = 110*S - 55*85 S^2 - 55*S + 85*S - 110*S = 0 S^2 - 80*S = 0 S*(S-80) = 0 ====> The only positive root is S = 80. Thus we found the distance between the homes: it is 80 meters. Having this known, you van calculate the ratio of their rates from (3) {{{V[1]/V[2]}}} = {{{(80-55)/55}}} = {{{25/55}}} = {{{5/11}}} and hence calculate the distance "x" from the Bob's home to the abandoned car from the proportion {{{20/x}}} = {{{5/11}}} ======> x = {{{(20*11)/5}}} = 44 m. 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