Question

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]

Answer 1

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?

Answer 2

oops, the equation is s=1+2t+1/4 t^2 in the original problem

Answer 3

So the equation is:
\[s(t)=1+2t+\frac{1}{4}t^2\]
?

Answer 4

Yes, that's the equation

Answer 5

I found V= S'(t) = 2t + (1/4) t

Answer 6

Well, maybe that's the issue, The derivative of \(s(t)\) Is:
\[s'(t)=2+\frac{t}{2}\]

Answer 7

Okay I'll check my algebra... but from there, how would I find the avg velocity

Answer 8

for [1,3]

Answer 9

Hmm. well isn't average value of a equation in the interval [a,b]:
\[\frac{f(b)+f(a)}{2}\]

Answer 10

okay.. so it's just (1+3)/2... the derivative doesn't even matter for this part?

Answer 11

So average velocity would be:
\[\frac{(2+0.5(3))+(2+0.5(1))}{2}=\frac{3.5+2.5}{2}=\frac{6}{2}=3\]

Answer 12

No you need to plug the 1 and 3 into the function \(s'(t)\) Like so:
\[Average=\frac{s'(3)+s'(1)}{2}\]

Answer 13

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?

Answer 14

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

Answer 15

alright, thanks!