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A particle moves according to the equation: `f(x,t) = 27sin(2pi(t/6.3 - x/0.21))` At...

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blehens | eNoter

Posted August 26, 2013 at 12:37 PM via web

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A particle moves according to the equation:

`f(x,t) = 27sin(2pi(t/6.3 - x/0.21))`

At what time will the particle be at co-ordinates (5, -6.29)?

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mathsworkmusic | (Level 3) Associate Educator

Posted August 26, 2013 at 1:52 PM (Answer #1)

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At the coordinates (x,y) = (5,-6.29), x = 5.

When x = 5

`f(x) = 27sin(2pi(t/6.3 - 5/0.21)) = 27sin(2pi(t/6.3 - 23.81)) `

`= 27sin(2pi((t-150)/6.3))`

and we are given that `f(x) = y = -6.29` so that

`sin(2pi((t-150)/6.3)) = -6.29/27 = -0.233`

Now, `sin^(-1)(-2.33) = -0.235` so that

`2pi((t-150)/6.3) = -0.235`  `implies`

`(t-150)/6.3 = -0.470pi `   `implies`

`t = -2.963pi + 150 = 141 `

Answer: the particle will be at (x,y) = (5,-6.29) at time t = 141

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