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Question Number 13449 by Tinkutara last updated on 20/May/17
Four particles A, B, C and D are situated at the corners of a square ABCD of side a at t = 0. Each of the particles moves with constant speed v. A always has its velocity along AB, B along BC, C along CD and D along DA. At what time will these particles meet each other?
$$\mathrm{Four}\:\mathrm{particles}\:{A},\:{B},\:{C}\:\mathrm{and}\:{D}\:\mathrm{are}\:\mathrm{situated} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{corners}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square}\:{ABCD}\:\mathrm{of}\:\mathrm{side} \\ $$$${a}\:\mathrm{at}\:{t}\:=\:\mathrm{0}.\:\mathrm{Each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particles}\:\mathrm{moves} \\ $$$$\mathrm{with}\:\mathrm{constant}\:\mathrm{speed}\:{v}.\:{A}\:\mathrm{always}\:\mathrm{has}\:\mathrm{its} \\ $$$$\mathrm{velocity}\:\mathrm{along}\:{AB},\:{B}\:\mathrm{along}\:{BC},\:{C}\:\mathrm{along} \\ $$$${CD}\:\mathrm{and}\:{D}\:\mathrm{along}\:{DA}.\:\mathrm{At}\:\mathrm{what}\:\mathrm{time}\:\mathrm{will} \\ $$$$\mathrm{these}\:\mathrm{particles}\:\mathrm{meet}\:\mathrm{each}\:\mathrm{other}? \\ $$
Answered by mrW1 last updated on 20/May/17
When the particles move, the distance between them will be reduced from a to 0 with the speed v. The time they need is t=(a/v)
$${When}\:{the}\:{particles}\:{move},\:{the}\:{distance} \\ $$$${between}\:{them}\:{will}\:{be}\:{reduced}\:{from}\:{a} \\ $$$${to}\:\mathrm{0}\:{with}\:{the}\:{speed}\:{v}.\:{The}\:{time}\:{they} \\ $$$${need}\:{is} \\ $$$${t}=\frac{{a}}{{v}} \\ $$
Commented by Tinkutara last updated on 20/May/17
But can you explain why they will meet? Since they all are moving with constant speeds in the same direction along the sides of a square, they should never meet.
$$\mathrm{But}\:\mathrm{can}\:\mathrm{you}\:\mathrm{explain}\:\mathrm{why}\:\mathrm{they}\:\mathrm{will}\:\mathrm{meet}? \\ $$$$\mathrm{Since}\:\mathrm{they}\:\mathrm{all}\:\mathrm{are}\:\mathrm{moving}\:\mathrm{with}\:\mathrm{constant} \\ $$$$\mathrm{speeds}\:\mathrm{in}\:\mathrm{the}\:\mathrm{same}\:\mathrm{direction}\:\mathrm{along}\:\mathrm{the} \\ $$$$\mathrm{sides}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square},\:\mathrm{they}\:\mathrm{should}\:\mathrm{never} \\ $$$$\mathrm{meet}. \\ $$
Commented by ajfour last updated on 20/May/17
this is simple and good.
$${this}\:{is}\:{simple}\:{and}\:{good}. \\ $$
Commented by ajfour last updated on 20/May/17
the square ABCD remains a square but decreases in edge length and turns about its centre where the vertices come together, in a time required for the edge length to vanish, at a rate=v. hence t=a/v .
$${the}\:{square}\:{ABCD}\:\:{remains}\:{a} \\ $$$${square}\:{but}\:{decreases}\:{in}\:{edge}\:{length} \\ $$$${and}\:{turns}\:{about}\:{its}\:{centre}\:{where} \\ $$$${the}\:{vertices}\:{come}\:{together},\:{in}\:{a}\: \\ $$$${time}\:{required}\:{for}\:{the}\:{edge}\:{length} \\ $$$${to}\:{vanish},\:{at}\:{a}\:{rate}=\boldsymbol{{v}}. \\ $$$${hence}\:{t}={a}/{v}\:. \\ $$
Commented by mrW1 last updated on 20/May/17
The track of each particle is curve which has the length a. At every time the position of the particles is a squar which rotates and gets smaller and smaller till a single point.
$${The}\:{track}\:{of}\:{each}\:{particle}\:{is}\:{curve}\:{which} \\ $$$${has}\:{the}\:{length}\:{a}.\:{At}\:{every}\:{time}\:{the} \\ $$$${position}\:{of}\:{the}\:{particles}\:{is}\:{a}\:{squar}\: \\ $$$${which}\:{rotates}\:{and}\:{gets}\:{smaller}\:{and} \\ $$$${smaller}\:{till}\:{a}\:{single}\:{point}. \\ $$
Commented by mrW1 last updated on 20/May/17
Commented by Tinkutara last updated on 20/May/17
Thanks to both mrW1 and ajfour.
$$\mathrm{Thanks}\:\mathrm{to}\:\mathrm{both}\:\mathrm{mrW1}\:\mathrm{and}\:\mathrm{ajfour}. \\ $$
Commented by mrW1 last updated on 20/May/17
I found some pictures in Internet
$${I}\:{found}\:{some}\:{pictures}\:{in}\:{Internet}\: \\ $$
Commented by mrW1 last updated on 20/May/17
Commented by mrW1 last updated on 20/May/17
Answered by ajfour last updated on 20/May/17
component of velocity towards centre =(v/( (√2))) distance to centre =(a/( (√2))) they will meet at the centre in a time t=(((a/(√2)))/((v/(√2)))) =(v/a) .
$${component}\:{of}\:{velocity}\:{towards} \\ $$$${centre}\:=\frac{{v}}{\:\sqrt{\mathrm{2}}}\: \\ $$$${distance}\:{to}\:{centre}\:=\frac{{a}}{\:\sqrt{\mathrm{2}}}\: \\ $$$${they}\:{will}\:{meet}\:{at}\:{the}\:{centre} \\ $$$${in}\:{a}\:{time}\:\boldsymbol{{t}}=\frac{\left(\boldsymbol{{a}}/\sqrt{\mathrm{2}}\right)}{\left(\boldsymbol{{v}}/\sqrt{\mathrm{2}}\right)}\:=\frac{\boldsymbol{{v}}}{\boldsymbol{{a}}}\:. \\ $$
Commented by ajfour last updated on 20/May/17

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