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anonymous

  • 4 years ago

s(t) = -16 t^2 + 80 t The rocket is traveling down at a velocity of 60 fps at what time?

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  1. anonymous
    • 4 years ago
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    calculus

  2. anonymous
    • 4 years ago
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    s'(t)=v(t)=-32t+80

  3. anonymous
    • 4 years ago
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    so 60=-32t+80 32t=20 t=20/32 s

  4. anonymous
    • 4 years ago
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    t=20/32 is the answer?

  5. anonymous
    • 4 years ago
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    thats wrong

  6. anonymous
    • 4 years ago
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    well 20/32 is the fraction, 0.625 seconds is what I got,

  7. anonymous
    • 4 years ago
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    I didn't have a calculator on hand, sorry if that's also wrong

  8. anonymous
    • 4 years ago
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    i know but I plugged it and it said it was wrong

  9. anonymous
    • 4 years ago
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    are you sure you posted the question correctly, or is there any other information, or unit conversion to be done?

  10. anonymous
    • 4 years ago
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    if not, i'm sorry I don't think i can help

  11. anonymous
    • 4 years ago
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    heres everything i know:

  12. anonymous
    • 4 years ago
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    The height of a rocket launched from the ground is given by the function s(t) = -16 t^2 + 80 t, where s is in feet above ground and t is in seconds. After 3 seconds, the rocket is 96 feet above ground. After 3 seconds, the rocket is traveling at a velocity of -16 feet per second. The rocket is traveling down at a velocity of 60 fps at what time? 0.625 Incorrect: Your answer is incorrect.

  13. anonymous
    • 4 years ago
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    ah, okay hold on here

  14. anonymous
    • 4 years ago
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    -60=-32t+80 32t=140 t=140/32 t=4.375 seconds

  15. anonymous
    • 4 years ago
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    the time i gave you originally was, the time it took to go from the top of it's flight (0fps) to 60fps in the downward direction, so not enough information was given originally.

  16. anonymous
    • 4 years ago
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    that should be correct now

  17. anonymous
    • 4 years ago
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    ok lets see....

  18. anonymous
    • 4 years ago
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    yeah its right thank you very much

  19. anonymous
    • 4 years ago
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    so how did you find it?

  20. anonymous
    • 4 years ago
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    so I can understand for another question

  21. anonymous
    • 4 years ago
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    well you know that v(t) = s'(t) and you know how to do calculus right?

  22. anonymous
    • 4 years ago
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    deriv of t = v

  23. anonymous
    • 4 years ago
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    not quite.. ds/dt=v(t)

  24. anonymous
    • 4 years ago
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    so the deriv or ds/dt = the velocity of t

  25. anonymous
    • 4 years ago
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    so the derivative of s with respect to t is the velocity in terms of t ..in words..

  26. anonymous
    • 4 years ago
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    ok

  27. anonymous
    • 4 years ago
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    kinda the same thing i said

  28. anonymous
    • 4 years ago
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    basically if you have s(t) and you want velocity you will do s'(t) by calculus methods which you should know how to do, if you're taking this class. then the acceleration function is v'(t) or s''(t) if you need to find that

  29. anonymous
    • 4 years ago
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    in this question since they said the velocity was -60 you can put that in the equation where v(t) is because -60 fps is the v(t) just like the acceleration was -32fps or v'(t)

  30. anonymous
    • 4 years ago
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    whats the difference between v prime of (t) and the second deriv of (t)

  31. anonymous
    • 4 years ago
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    which is gravity in america.. aha. crazy americans.

  32. anonymous
    • 4 years ago
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    you would never say the second derivative of (t) it would be worded as the second derivative of s(t) or s''(t) or s double prime of t, and that is the acceleration function of t so s''(t) = v'(t) = a(t)

  33. anonymous
    • 4 years ago
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    oh ok

  34. anonymous
    • 4 years ago
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    hey I just started this class so we havent gotten that far

  35. anonymous
    • 4 years ago
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    t is always what the function is with respect to, it is the variable you are measuring, you are not deriving it, you are deriving in terms of it

  36. anonymous
    • 4 years ago
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    and that's fine! calculus isn't an easy concept to grasp but when you do.. boy i tell ya. good times.. aha.

  37. anonymous
    • 4 years ago
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    lol

  38. anonymous
    • 4 years ago
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    is this highschool?

  39. anonymous
    • 4 years ago
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    or university

  40. anonymous
    • 4 years ago
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    no college im attending u of maryland

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