The position of an object at time t is given by s(t) = -8 - 9t. Find the instantaneous velocity at t = 1 by finding the derivative. I don't get this in word problem form
what's the derivative of the given s(t) with respect to t?
s(t) represents the distance in terms of variable t. instantaneous velocity ( velocity ) is - as we know - is the rate of change of distance . In math terms - rate of change is given by derivative. so we need to have derivative of s(t) to have velocity. instantaneous velocity is the velocity at a particular instant , here he's asking for velocity at t = 1 :)
find the derivative of s(t). substitute t = 1 in the result and there you will have answer to the question.
so do you just plug in 1 for x? I'm confused since there's no exponent
I just dont get how to find the derivative since theres no exponent because I only knew how to do the power rule
when its given \[x^2\] you know its exponent is 2 right? that's how when it's just x. you consider it \[x^1\] -> exponent is 1. :)
OH IM DUMB, thank you! Let me do this real quick
no problem, it happens at times :)
Okay, so since it's 1, nothing really changes, right? I got -17 as the answer but I know thats wrong.
derivative of constant is 0. FYI
Ugh, I'm really sorry but I don't understand that.. Could you explain?
do you know this definition: \[\lim_{h \rightarrow 0}\frac{s(t+h)-s(t)}{h}=s'(t)\]
I don't know how to use it, no
actually you have a line just fine the slope of your line
what is the slope of s(t)=-9t-8?
for example the slope of y=mx+b is m so the slope of y=-9x-8 is ?
hmm, derivative with respect to t means how equation changes with respect to variable t. so when we say \[t^1\] now for t= 1 -> we have t^1 values = 1 for t =2 -> we havr t^1 values = 2 for t =3 -> we have t^1 values = 3 so change is 2-1 = 3-2 = 1{ constant value }
-9
that is your answer
derivative=slope
really?? That's all it is?
yep
@SandeepReddy , oh I understand now!
Thank you both so much!!!(:
pleasure is ours :)
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