anonymous
  • anonymous
s(t) = -16 t^2 + 80 t The rocket is traveling down at a velocity of 60 fps at what time?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
calculus
anonymous
  • anonymous
s'(t)=v(t)=-32t+80
anonymous
  • anonymous
so 60=-32t+80 32t=20 t=20/32 s

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anonymous
  • anonymous
t=20/32 is the answer?
anonymous
  • anonymous
thats wrong
anonymous
  • anonymous
well 20/32 is the fraction, 0.625 seconds is what I got,
anonymous
  • anonymous
I didn't have a calculator on hand, sorry if that's also wrong
anonymous
  • anonymous
i know but I plugged it and it said it was wrong
anonymous
  • anonymous
are you sure you posted the question correctly, or is there any other information, or unit conversion to be done?
anonymous
  • anonymous
if not, i'm sorry I don't think i can help
anonymous
  • anonymous
heres everything i know:
anonymous
  • anonymous
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.
anonymous
  • anonymous
ah, okay hold on here
anonymous
  • anonymous
-60=-32t+80 32t=140 t=140/32 t=4.375 seconds
anonymous
  • anonymous
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.
anonymous
  • anonymous
that should be correct now
anonymous
  • anonymous
ok lets see....
anonymous
  • anonymous
yeah its right thank you very much
anonymous
  • anonymous
so how did you find it?
anonymous
  • anonymous
so I can understand for another question
anonymous
  • anonymous
well you know that v(t) = s'(t) and you know how to do calculus right?
anonymous
  • anonymous
deriv of t = v
anonymous
  • anonymous
not quite.. ds/dt=v(t)
anonymous
  • anonymous
so the deriv or ds/dt = the velocity of t
anonymous
  • anonymous
so the derivative of s with respect to t is the velocity in terms of t ..in words..
anonymous
  • anonymous
ok
anonymous
  • anonymous
kinda the same thing i said
anonymous
  • anonymous
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
anonymous
  • anonymous
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)
anonymous
  • anonymous
whats the difference between v prime of (t) and the second deriv of (t)
anonymous
  • anonymous
which is gravity in america.. aha. crazy americans.
anonymous
  • anonymous
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)
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
hey I just started this class so we havent gotten that far
anonymous
  • anonymous
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
anonymous
  • anonymous
and that's fine! calculus isn't an easy concept to grasp but when you do.. boy i tell ya. good times.. aha.
anonymous
  • anonymous
lol
anonymous
  • anonymous
is this highschool?
anonymous
  • anonymous
or university
anonymous
  • anonymous
no college im attending u of maryland

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