anonymous
  • anonymous
The displacement (in meters) of an object moving in a straight line is given by s=1+2t+1/4t^2^2, where t is measured in seconds. Find the avg velocity over each time period, a) [1,3]
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
I solved the problem by saying that velocity would be the derivitave of s(t), and found that v= 2t +1/4t..... then I plugged in 1 for t in my equation but I'm getting 9/4 which is wrong.... the answer is supposed to = 3. Any ideas how to solve this?
anonymous
  • anonymous
oops, the equation is s=1+2t+1/4 t^2 in the original problem
anonymous
  • anonymous
So the equation is: \[s(t)=1+2t+\frac{1}{4}t^2\] ?

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anonymous
  • anonymous
Yes, that's the equation
anonymous
  • anonymous
I found V= S'(t) = 2t + (1/4) t
anonymous
  • anonymous
Well, maybe that's the issue, The derivative of \(s(t)\) Is: \[s'(t)=2+\frac{t}{2}\]
anonymous
  • anonymous
Okay I'll check my algebra... but from there, how would I find the avg velocity
anonymous
  • anonymous
for [1,3]
anonymous
  • anonymous
Hmm. well isn't average value of a equation in the interval [a,b]: \[\frac{f(b)+f(a)}{2}\]
anonymous
  • anonymous
okay.. so it's just (1+3)/2... the derivative doesn't even matter for this part?
anonymous
  • anonymous
So average velocity would be: \[\frac{(2+0.5(3))+(2+0.5(1))}{2}=\frac{3.5+2.5}{2}=\frac{6}{2}=3\]
anonymous
  • anonymous
No you need to plug the 1 and 3 into the function \(s'(t)\) Like so: \[Average=\frac{s'(3)+s'(1)}{2}\]
anonymous
  • anonymous
okay that makes sense. Now if I need to find the instantaneous velocity when t=1... then I would plug 1 in so v=2t + (t/2) V=2(1) + (1/2) V=2 1/2?
anonymous
  • anonymous
Well \(s'(1)=2\frac{1}{2}\) But s'(t)=\(2+\frac{t}{2}\) Remember haha its those small things that make a big difference
anonymous
  • anonymous
alright, thanks!

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