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Question Number 182405 by abdullah_ff last updated on 09/Dec/22

Zombie apocalypse has started. You  are at point A (3,2). At point B (−5,4)  there is a shelter. A river flows along   the X axis. You are running at a constant  velocity 1 ms^(−1)  with an intention to  reach the shelter but before that, you  have to reach the river and fill-up your  water jar. What is the lowest time it   takes to reach B from A? [It takes 1 second  time to fill the jar with water.]     a) 11.0 s  b) 8.24 s  c) 10.1 s  d) 9 s

$${Zombie}\:{apocalypse}\:{has}\:{started}.\:{You} \\ $$$${are}\:{at}\:{point}\:{A}\:\left(\mathrm{3},\mathrm{2}\right).\:{At}\:{point}\:{B}\:\left(−\mathrm{5},\mathrm{4}\right) \\ $$$${there}\:{is}\:{a}\:{shelter}.\:{A}\:{river}\:{flows}\:{along}\: \\ $$$${the}\:{X}\:{axis}.\:{You}\:{are}\:{running}\:{at}\:{a}\:{constant} \\ $$$${velocity}\:\mathrm{1}\:{ms}^{−\mathrm{1}} \:{with}\:{an}\:{intention}\:{to} \\ $$$${reach}\:{the}\:{shelter}\:{but}\:{before}\:{that},\:{you} \\ $$$${have}\:{to}\:{reach}\:{the}\:{river}\:{and}\:{fill}-{up}\:{your} \\ $$$${water}\:{jar}.\:{What}\:{is}\:{the}\:{lowest}\:{time}\:{it}\: \\ $$$${takes}\:{to}\:{reach}\:{B}\:{from}\:{A}?\:\left[{It}\:{takes}\:\mathrm{1}\:{second}\right. \\ $$$$\left.{time}\:{to}\:{fill}\:{the}\:{jar}\:{with}\:{water}.\right]\: \\ $$$$ \\ $$$$\left.{a}\right)\:\mathrm{11}.\mathrm{0}\:{s} \\ $$$$\left.{b}\right)\:\mathrm{8}.\mathrm{24}\:{s} \\ $$$$\left.{c}\right)\:\mathrm{10}.\mathrm{1}\:{s} \\ $$$$\left.{d}\right)\:\mathrm{9}\:{s} \\ $$

Commented by mr W last updated on 09/Dec/22

the question would be more interesting,  when you with filled water jar only   run at a speed 0.6 ms^(−1) .

$${the}\:{question}\:{would}\:{be}\:{more}\:{interesting}, \\ $$$${when}\:{you}\:{with}\:{filled}\:{water}\:{jar}\:{only}\: \\ $$$${run}\:{at}\:{a}\:{speed}\:\mathrm{0}.\mathrm{6}\:{ms}^{−\mathrm{1}} . \\ $$

Answered by CrispyXYZ last updated on 09/Dec/22

a     This question can be converted to:  A(3, 2),  B(−5, 4), P(p, 0)  Find t_(min) =(AP+BP+1)_(min)    Algebra method:  AP=(√((p−3)^2 +2^2 ))  BP=(√((p+5)^2 +4^2 ))  t≥(√((p−3+p+5)^2 +6^2 ))+1=11  t_(min) =11

$${a} \\ $$$$\: \\ $$$$\mathrm{This}\:\mathrm{question}\:\mathrm{can}\:\mathrm{be}\:\mathrm{converted}\:\mathrm{to}: \\ $$$${A}\left(\mathrm{3},\:\mathrm{2}\right),\:\:{B}\left(−\mathrm{5},\:\mathrm{4}\right),\:{P}\left({p},\:\mathrm{0}\right) \\ $$$$\mathrm{Find}\:{t}_{\mathrm{min}} =\left({AP}+{BP}+\mathrm{1}\right)_{\mathrm{min}} \\ $$$$\:\mathrm{Algebra}\:\mathrm{method}: \\ $$$${AP}=\sqrt{\left({p}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} } \\ $$$${BP}=\sqrt{\left({p}+\mathrm{5}\right)^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} } \\ $$$${t}\geqslant\sqrt{\left({p}−\mathrm{3}+{p}+\mathrm{5}\right)^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} }+\mathrm{1}=\mathrm{11} \\ $$$${t}_{\mathrm{min}} =\mathrm{11} \\ $$

Commented by CrispyXYZ last updated on 09/Dec/22

Commented by abdullah_ff last updated on 09/Dec/22

thank you

$${thank}\:{you} \\ $$

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