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Mathematics 23 Online
OpenStudy (anonymous):

If h(t) represents the height of an object above ground level at time t and h(t) is given by: h(t)=-16t^2+14t+1 find the speed at time t=0.

OpenStudy (anonymous):

this is physics. use that forum.

OpenStudy (anonymous):

im learning this in calculus

OpenStudy (anonymous):

oh cool. calc AB or BC. that determines how to do this.

OpenStudy (anonymous):

idk. its my first calc class. its the intro course.

OpenStudy (anonymous):

uhhh... do you know how to differentiate? or take limits?

OpenStudy (anonymous):

we just started doing that in class. not too keen on it yet.

OpenStudy (anonymous):

ok so the derivative of position is velocity because the "change" in position is velocity. so differentiate h(x)

OpenStudy (anonymous):

i dont know what it means to differentiate

OpenStudy (anonymous):

i thought you just plug in for T?

OpenStudy (anonymous):

ok do you know how to: 1) take a derivative 2) find a limit? which one do you know...basically what are you doing in class now?

OpenStudy (anonymous):

we just started derivatives and limits... we've mainly been doing rate of change where you plug in something in the function.

OpenStudy (anonymous):

eh. you have to take the derivative of that.. or just use the limit definition of the derivative to take the derivative...you need to do one.

OpenStudy (anonymous):

can you show me how to do that?

OpenStudy (anonymous):

v(t) = dh/dt = -32t + 14

OpenStudy (anonymous):

v(0) = 0 + 14

OpenStudy (anonymous):

Therefore , Answer = 14

OpenStudy (anonymous):

ok so the derivative of something is taking the 2 from the square and multiplying it by the coefficient... what happened to the +1 at the end of the function?

OpenStudy (phi):

derivative of a constant = 0

OpenStudy (phi):

think of it as 1*t^0

OpenStudy (anonymous):

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