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

- anonymous

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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- anonymous

calculus

- anonymous

s'(t)=v(t)=-32t+80

- anonymous

so 60=-32t+80
32t=20
t=20/32 s

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## More answers

- anonymous

t=20/32 is the answer?

- anonymous

thats wrong

- anonymous

well 20/32 is the fraction, 0.625 seconds is what I got,

- anonymous

I didn't have a calculator on hand, sorry if that's also wrong

- anonymous

i know but I plugged it and it said it was wrong

- anonymous

are you sure you posted the question correctly, or is there any other information, or unit conversion to be done?

- anonymous

if not, i'm sorry I don't think i can help

- anonymous

heres everything i know:

- 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

ah, okay hold on here

- anonymous

-60=-32t+80
32t=140
t=140/32
t=4.375 seconds

- 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

that should be correct now

- anonymous

ok lets see....

- anonymous

yeah its right thank you very much

- anonymous

so how did you find it?

- anonymous

so I can understand for another question

- anonymous

well you know that v(t) = s'(t)
and you know how to do calculus right?

- anonymous

deriv of t = v

- anonymous

not quite..
ds/dt=v(t)

- anonymous

so the deriv or ds/dt = the velocity of t

- anonymous

so the derivative of s with respect to t is the velocity in terms of t
..in words..

- anonymous

ok

- anonymous

kinda the same thing i said

- 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

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

whats the difference between v prime of (t) and the second deriv of (t)

- anonymous

which is gravity in america.. aha. crazy americans.

- 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

oh ok

- anonymous

hey I just started this class so we havent gotten that far

- 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

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

- anonymous

lol

- anonymous

is this highschool?

- anonymous

or university

- anonymous

no college im attending u of maryland

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