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Three-towns-X-Y-and-Z-are-on-a-straight-road-and-Y-is-the-mid-way-between-X-and-Z-A-motor-cyclist-moving-with-uniform-acceleration-passes-X-Y-and-Z-The-speed-with-which-the-motocyclist-passes-X-a




Question Number 26328 by tawa tawa last updated on 24/Dec/17
Three towns X, Y and Z are on a straight road and Y is the mid−way between  X and Z. A motor cyclist moving with uniform acceleration passes X, Y and Z.   The speed with which the motocyclist passes X and Z are 20m/s and 40m/s   respectively. Find the speed with which the motorcyclist passes Y.
$$\mathrm{Three}\:\mathrm{towns}\:\mathrm{X},\:\mathrm{Y}\:\mathrm{and}\:\mathrm{Z}\:\mathrm{are}\:\mathrm{on}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{road}\:\mathrm{and}\:\mathrm{Y}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mid}−\mathrm{way}\:\mathrm{between} \\ $$$$\mathrm{X}\:\mathrm{and}\:\mathrm{Z}.\:\mathrm{A}\:\mathrm{motor}\:\mathrm{cyclist}\:\mathrm{moving}\:\mathrm{with}\:\mathrm{uniform}\:\mathrm{acceleration}\:\mathrm{passes}\:\mathrm{X},\:\mathrm{Y}\:\mathrm{and}\:\mathrm{Z}.\: \\ $$$$\mathrm{The}\:\mathrm{speed}\:\mathrm{with}\:\mathrm{which}\:\mathrm{the}\:\mathrm{motocyclist}\:\mathrm{passes}\:\mathrm{X}\:\mathrm{and}\:\mathrm{Z}\:\mathrm{are}\:\mathrm{20m}/\mathrm{s}\:\mathrm{and}\:\mathrm{40m}/\mathrm{s}\: \\ $$$$\mathrm{respectively}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{with}\:\mathrm{which}\:\mathrm{the}\:\mathrm{motorcyclist}\:\mathrm{passes}\:\mathrm{Y}. \\ $$
Commented by tawa tawa last updated on 24/Dec/17
please help.
$$\mathrm{please}\:\mathrm{help}. \\ $$
Answered by jota@ last updated on 25/Dec/17
  s_Z −s_X =2(s_Y −s_X ).    v_Z ^2 −v_X ^2 =2a(s_Z −s_X )  1200=2a(s_Z −s_X )=4a(s_Y −s_X )  600=2a(s_Y −s_X ).  v_Y ^2 =v_X ^2 +2a(s_Y −s_X )  v_Y ^2 =400+600  v_Y =10(√(10))
$$ \\ $$$${s}_{{Z}} −{s}_{{X}} =\mathrm{2}\left({s}_{{Y}} −{s}_{{X}} \right). \\ $$$$ \\ $$$${v}_{{Z}} ^{\mathrm{2}} −{v}_{{X}} ^{\mathrm{2}} =\mathrm{2}{a}\left({s}_{{Z}} −{s}_{{X}} \right) \\ $$$$\mathrm{1200}=\mathrm{2}{a}\left({s}_{{Z}} −{s}_{{X}} \right)=\mathrm{4}{a}\left({s}_{{Y}} −{s}_{{X}} \right) \\ $$$$\mathrm{600}=\mathrm{2}{a}\left({s}_{{Y}} −{s}_{{X}} \right). \\ $$$${v}_{{Y}} ^{\mathrm{2}} ={v}_{{X}} ^{\mathrm{2}} +\mathrm{2}{a}\left({s}_{{Y}} −{s}_{{X}} \right) \\ $$$${v}_{{Y}} ^{\mathrm{2}} =\mathrm{400}+\mathrm{600} \\ $$$${v}_{{Y}} =\mathrm{10}\sqrt{\mathrm{10}} \\ $$
Commented by tawa tawa last updated on 25/Dec/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\: \\ $$

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