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.
this is physics. use that forum.
im learning this in calculus
oh cool. calc AB or BC. that determines how to do this.
idk. its my first calc class. its the intro course.
uhhh... do you know how to differentiate? or take limits?
we just started doing that in class. not too keen on it yet.
ok so the derivative of position is velocity because the "change" in position is velocity. so differentiate h(x)
i dont know what it means to differentiate
i thought you just plug in for T?
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?
we just started derivatives and limits... we've mainly been doing rate of change where you plug in something in the function.
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.
can you show me how to do that?
v(t) = dh/dt = -32t + 14
v(0) = 0 + 14
Therefore , Answer = 14
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?
derivative of a constant = 0
think of it as 1*t^0
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