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v.s

  • 3 years ago

(b) Find the position, velocity, and acceleration of the mass at time t = 5π/6. x(t) = 8 cos t v(t) = -8sint a(t) = -8cost x(5π/6)= v(5π/6)= a(5π/6) =

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  1. v.s
    • 3 years ago
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    i get decimals for the answers, i'm just wonder how i would answer it on a test

  2. Coolsector
    • 3 years ago
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    cos(5pi/6) = -sqrt(3)/2 sin(5pi/6) = 0.5

  3. Coolsector
    • 3 years ago
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    helps ?

  4. phi
    • 3 years ago
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    you should memorize sin,cos and tan of 0,30,45, 60 and 90 degrees you should know how to change between radians and degrees radians *180/pi or degrees*pi/180 you should know how to figure out the signs depending on which quadrant you are in for sin, cos, tan

  5. cwrw238
    • 3 years ago
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    a = -8 cos (5pi/6)

  6. v.s
    • 3 years ago
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    okaay so i have to answer in radiants

  7. phi
    • 3 years ago
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    if I were doing this, I would say 5pi/6 is the same as 30 degrees in the 2nd quadrant. sin will be +, cos -

  8. v.s
    • 3 years ago
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    x(5pi/6)=-4sqroot3

  9. qpHalcy0n
    • 3 years ago
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    That is correct for 8cos(5pi / 6)

  10. v.s
    • 3 years ago
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    okaay

  11. qpHalcy0n
    • 3 years ago
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    So you have the concept, just apply it to the rest.

  12. v.s
    • 3 years ago
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    v(t)=-4

  13. v.s
    • 3 years ago
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    a(t)=4sqroot3

  14. Coolsector
    • 3 years ago
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    v(5π/6)=-4 .. right

  15. Coolsector
    • 3 years ago
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    a(5pi/6) right as well

  16. v.s
    • 3 years ago
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    thank you guys learned alot

  17. phi
    • 3 years ago
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    I don't know that it really helps me memorize sin of these angles, it is interesting: angle sin(angle) 0 \(\frac{\sqrt{0}}{2}\) 30 \(\frac{\sqrt{1}}{2}\) 45 \(\frac{\sqrt{2}}{2}\) 60 \(\frac{\sqrt{3}}{2}\) 90 \(\frac{\sqrt{4}}{2}\)

  18. v.s
    • 3 years ago
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    thank you

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