The velocity function of a moving particle on a coordinate line is v(t) = 3 cos(2t) for 0 is less than or equal to t and t is greater than or equal to 2pi (c) Determine the distance travelled by the particle during 0 is less than or equal to t is greater than or equal to 2pi.
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If you need to find the displacement of moving particle, under the given conditions, you need to evaluate the definite integral of the given function such that:
`s(t) = int_0^(2pi) 3cos 2t dt `
You need to come up with the following substitution such that:
`2t = u => 2dt = du => dt = (du)/2`
`{(t = 0 => u = 0 ),(t = 2pi => u = 4pi ):}`
Changing the variable yields:
`int_0^(2pi) 3cos 2t dt = int_0^(4pi) 3cosu(du)/2`
`(3/2) int_0^(4pi) cos u du = (3/2) sin u|_0^(4pi)`
Using the fundamental theorem of calculus yields:
`(3/2) int_0^(4pi) cos u du = (3/2)(sin 4pi - sin 0) = 0`
Hence, evaluating the distance travelled by the moving particle yields `s(t) = 0.`
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