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Thread: trig help please

  1. #1
    Apr 2009

    trig help please

    i have two questions and i also have the answers to them, but i dont know how they gotten them. can someone please explain it step by step?

    1. the productivity of a person at work (on a scale of 0 to 10) is modelled by a cosine function: 5cos((pi)t/2) + 5, where t is in hours. If the person starts work at t=0, being 8:00 am, what times is the worker the least productive?

    answer: 10a.m. and

    2.At t=0 a car is due 2km north of you heading west moving 13km/s after 1.5s. what is the angle in radians between you and the car from where the car was at t=0

    answer: 1.469radians

    i appreciate the help ! (:
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  2. #2
    Apr 2009

    Do you know how to plot the graph of 5cos((pi)t/2) + 5 ? You'll probably have to know about basic graph transformations to do so. If you're not sure about these, this is pretty useful, it explains all the basic rules: New Functions from Old

    And here's what it looks like:

    So the lowest productivity is where the graph has the lowest value. On here it's where the graph hits zero. Go along the x-axis counting up the hours, starting with 8am at 0, and you'll find 10am and 2pm are your answers
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  3. #3
    Apr 2009
    United Kingdom

    Question 2

    As for your second question, what you need to do can be found by drawing a couple of diagrams.

    Consider what the situation looks like at t = 0 seconds. The car is 2km due north of you:

    Next, consider what happens between 0 seconds and 1.5 seconds. The car moves due west at 13km/s for 1.5 seconds. Using distance = speed x time, this means that the car moves 13 x 1.5 = 19.5km during these 1.5 seconds. We therefore have a situation like this:

    Note that we have a right-angle in the upper-right corner of the triangle since the car moved due west. We need to calculate \theta.

    So, using trigonometry,

    \begin{array}{l}<br />
 \tan \theta  = \frac{{{\rm{opposite}}}}{{{\rm{adjacent}}}} \\ <br />
 \tan \theta  = \frac{{19.5}}{2} \\ <br />
 \theta  = \arctan \left( {\frac{{19.5}}{2}} \right) \\ <br />
 \theta  = 1.469 \\ <br />

    to three decimal places.
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