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Mathematics 16 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

calculus

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

t=20/32 is the answer?

OpenStudy (anonymous):

thats wrong

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

heres everything i know:

OpenStudy (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.

OpenStudy (anonymous):

ah, okay hold on here

OpenStudy (anonymous):

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

OpenStudy (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.

OpenStudy (anonymous):

that should be correct now

OpenStudy (anonymous):

ok lets see....

OpenStudy (anonymous):

yeah its right thank you very much

OpenStudy (anonymous):

so how did you find it?

OpenStudy (anonymous):

so I can understand for another question

OpenStudy (anonymous):

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

OpenStudy (anonymous):

deriv of t = v

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

kinda the same thing i said

OpenStudy (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

OpenStudy (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)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (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)

OpenStudy (anonymous):

oh ok

OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

lol

OpenStudy (anonymous):

is this highschool?

OpenStudy (anonymous):

or university

OpenStudy (anonymous):

no college im attending u of maryland

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