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Math Help - Help with trigonometry question

  1. #1
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    Help with trigonometry question

    Hi, i need help in this trignometry question and i don't quite understand how to solve it this question is dealing with the equation:

    y-k sin or cos (x-h)
    ---- = -----------------
    b a

    just in case your not sure

    k = vertical shift
    b = amplitude
    h = horizontal shift
    a = period

    here is the question:

    a tire with a diameter of 63 cm coasts at 1.6 m/s over a nail at exactly 12:00 noon. the nail is stuck in the tire but does not deflate

    a. how high above the ground will the nail be at:
    1. 12:00:08
    2. 12:00:15
    3: 12:01:00
    4: 12:15:00

    b. Find the first four times the nail at 10 cm above the ground

    thanks! i really need to understand this question
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  2. #2
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    Quote Originally Posted by lauangela View Post
    Hi, i need help in this trignometry question and i don't quite understand how to solve it this question is dealing with the equation:

    y-k sin or cos (x-h)
    ---- = -----------------
    b a

    just in case your not sure

    k = vertical shift
    b = amplitude
    h = horizontal shift
    a = period

    here is the question:

    a tire with a diameter of 63 cm coasts at 1.6 m/s over a nail at exactly 12:00 noon. the nail is stuck in the tire but does not deflate

    a. how high above the ground will the nail be at:
    1. 12:00:08
    2. 12:00:15
    3: 12:01:00
    4: 12:15:00

    b. Find the first four times the nail at 10 cm above the ground

    thanks! i really need to understand this question
    the first thing you need to determine is the period of up and down motion for the nail.

    \omega = \frac{v}{r} = \frac{160 \, cm/s}{31.5 \, cm} = \frac{320}{63} \, rad/s

    period, T = \frac{2\pi}{\omega} = \frac{63\pi}{160} \approx 1.24 \, s

    at t = 0, the nail is on the ground and rises to a height of 63 cm at t = \frac{63\pi}{320} \approx 0.62 \, s

    since the cycle starts at a low point (h = 0) , goes up to a height of h = 63 cm and back down, a negative cosine curve is probably the best model to simulate the motion of the nail up and down over time.

    h = -31.5 \cos\left(\frac{320}{63} \cdot t\right) + 31.5

    where h is height above the ground in cm and t is time in seconds.
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  3. #3
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    thanks for the reply but i am still very confuse about the whole thing

    what about the
    1. 12:00:08
    2. 12:00:15
    3: 12:01:00
    4: 12:15:00

    how high?

    im sorry but i'm just confuse by the whole work

    i need to identify the periods, amplitude, horizontal and vertical shift but then i can't see it and overall confuse
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  4. #4
    MHF Contributor
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    North Texas
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    Quote Originally Posted by lauangela View Post
    thanks for the reply but i am still very confuse about the whole thing

    what about the
    1. 12:00:08
    2. 12:00:15
    3: 12:01:00
    4: 12:15:00

    how high?

    im sorry but i'm just confuse by the whole work

    i need to identify the periods, amplitude, horizontal and vertical shift but then i can't see it and overall confuse

    If you cannot determine this information given the height equation and graph provided, then you need to have a talk with your instructor.
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