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Question Number 14939 by Tinkutara last updated on 05/Jun/17
A point moves in x-y plane according  to the law x = 4 sin 6t and  y = 4(1 − cos 6t). Find distance traversed  by the particle in 5 seconds, when x and  y are in metres.
$$\mathrm{A}\:\mathrm{point}\:\mathrm{moves}\:\mathrm{in}\:{x}-{y}\:\mathrm{plane}\:\mathrm{according} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{law}\:{x}\:=\:\mathrm{4}\:\mathrm{sin}\:\mathrm{6}{t}\:\mathrm{and} \\ $$$${y}\:=\:\mathrm{4}\left(\mathrm{1}\:−\:\mathrm{cos}\:\mathrm{6}{t}\right).\:\mathrm{Find}\:\mathrm{distance}\:\mathrm{traversed} \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{in}\:\mathrm{5}\:\mathrm{seconds},\:\mathrm{when}\:{x}\:\mathrm{and} \\ $$$${y}\:\mathrm{are}\:\mathrm{in}\:\mathrm{metres}. \\ $$
Answered by ajfour last updated on 05/Jun/17
x^2 +(y−4)^2 =4^2   particle moves in a circle of radius  4m with centre as (0, 4m).   v=ωr =(6rad/s)(4m) =24m/s.   distance traveled in 5s is   d=vt =(24m/s)(5s) =120m .
$${x}^{\mathrm{2}} +\left({y}−\mathrm{4}\right)^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} \\ $$$${particle}\:{moves}\:{in}\:{a}\:{circle}\:{of}\:{radius} \\ $$$$\mathrm{4}{m}\:{with}\:{centre}\:{as}\:\left(\mathrm{0},\:\mathrm{4}{m}\right). \\ $$$$\:{v}=\omega{r}\:=\left(\mathrm{6}{rad}/{s}\right)\left(\mathrm{4}{m}\right)\:=\mathrm{24}{m}/{s}. \\ $$$$\:{distance}\:{traveled}\:{in}\:\mathrm{5}{s}\:{is} \\ $$$$\:{d}={vt}\:=\left(\mathrm{24}{m}/{s}\right)\left(\mathrm{5}{s}\right)\:=\mathrm{120}{m}\:. \\ $$
Commented by Tinkutara last updated on 05/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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