Calculus Rectilinear Motion Question...s(t), v(t), and a(t)?

Question

I have this question on my math homework I can't seem to find the correct way to do the last part of it. The question reads "A particle moves in a straight line so that after t minutes, its distance is s(t)=2(t^3) + 3t +1 feet from the origin.

a: Find the average velocity over the interval [1,5]
b: Find the instantaneous velocity at t=4
c: Find the total distance traveled (nearest tenth)
d: Find the displacement (nearest tenth)"

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a: I think I got this part, do I use the msec=... formula?
b: Do I just take the derivative of the s(t) formula to make it v(t) and solve for v(4)?
c: I got an imaginary number for this part, so I know something's not right. I've looked over the worksheet, this question, and my work, and I did it right.
d: Where do I find the displacement from? I had a formula sheet, but I don't know where it is now.

Thanks again to all who help!

for part a i used msec= f(x1)-f(x0) / x1 - x0 and used the s(t) formula as my f(x), 5 as x1, and 1 as x0. I got 65 ft/sec for that one.

for part b, I got the same thing, so I have no problems turning that in like that.

parts c & d: sorry I wasn't more clear...the instructor wants the answers to the nearest tenth, but rational numbers would work too, he said.
there's no "total time in the equation. do I need to solve for t in my velocity equation to find it? the question I posted was the complete question out of the book.

Answer 1

Well, I think that for the first oneyou can get the acceleration by deriving it twice, then you get the integral from 1 to 5, let's see...

ds/dt= 6t^2+3=v
dv/dt= 12t=a

for a

I think you should do it this way

integral of vdt/ integral of dt from 1 to 5

so you will get

(2t^3+3t)/t from 1 to 5

so it will be

2t^2+3 from 1 to 5

evaluating you'll get

2*(25)+3-2-3

=50-2

average speed = 48

for b

you just substitute t=4 to the first derivative

V=6*(4)^2+3=99ft/min