Velocity = 3cos(t)-2sin(t)?

t stands for time:

V(t) = 3cos(t)-2sin(t)

When is the speed the fastest?

2 Answers

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  • 1 decade ago
    Favorite Answer

    Express V(t) in the form R cos (t+a)

    3 cos t - 2 sin t = R cos (t+a) = R cos t cos a - R sin t sin a

    3 cos t = R cos t cos a

    3 = R cos a

    2 sin t = R sin t sin a

    2 = R sin a

    R^2 (sin^2 a + cos^2 a) = 2^2 +3^2

    R = sqrt (13)

    sin a = 2 / sqrt(13) cos a = 3 / sqrt(13)

    a=0.588 radians

    Velocity = 3 cos (t) - 2 sin (t) = SQRT(13) * cos (t+0.588)

    Cos (x) has a maximum value of 1, occurring at x=0, 2pi, 4pi, ... and minimum value of -1, occurring at x= pi, 3pi, ...

    Therefore the speed will be at the maximum of +sqrt(13)

    at:

    2pi-0.588 = 5.70

    4pi-0.588 = 11.98

    6pi-0.588

    and so on for even multiples of pi.

    The speed will be at the minimum of -sqrt(13) at:

    3pi-0.588=8.84

    5pi-0.588=15.12

    7pi-0.588

    and so on for odd multiples of pi.

  • 1 decade ago

    V(t)

    = 3cos(t)-2sin(t)

    = √13[(3/√13)cos(t) - (2/√13)sin(t)]

    = -√13sin[t-arctan(3/2)]

    |v| (max) = √13

    Solve |v(t)| = √13,

    t = pi/2 + arctan(3/2) + n*pi = 2.5536 + n*pi, n can be any integer.

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    Reason:

    |v| (max) = the amplitude of v(t).

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    r_k_stan,

    Your answer is the same as mine.

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