# Displacement, velocity, acceleration

A particle moves along a line so that its position at any time t>0 is given by the funciton s(t) = t^3 - 8t + 1, where s is measured in feet and t is measured in seconds.
(a) Find the displacement during the first 3 seconds.
(b) Find the average velocity during the first 3 seconds.
(c) Find the instantaneous velocity when t = 3.
(d) Find the acceleration of the particle when t = 3.
(e) At what value or values of t does the particle change direction?

Ok, this is pretty much just to check my answers because i'm paranoid.. Here are the answers i came out with..

(a) 4 ft
(b) 1 ft/s
(c) 19 ft/s
(d) 18 ft/s^2
(e) sqrt(8/3)s or 1.633333s

Is this correct?

Thanks

LeonhardEuler
Gold Member
Jacobpm64 said:
(b) 1 ft/s
Check this one again. Other than that the answers seem fine.

LeonhardEuler said:
Check this one again. Other than that the answers seem fine.

Yea except for that one I think they're all right.

let's see.. for that one i did..

s(t) = t^3 - 8t + 1
s(0) = 0^3 - 8(0) + 1 = 1
s(3) = 3^3 - 8(3) + 1 = 4
so it would be
average velocity = change in displacement/change in time
average velocity = (4 - 1) / (3 - 0) = 3/3 = 1 ft/s

not sure on this? can you point me in the right direction?

Jacobpm64 said:
let's see.. for that one i did..
s(t) = t^3 - 8t + 1
s(0) = 0^3 - 8(0) + 1 = 1
s(3) = 3^3 - 8(3) + 1 = 4
so it would be
average velocity = change in displacement/change in time
average velocity = (4 - 1) / (3 - 0) = 3/3 = 1 ft/s
not sure on this? can you point me in the right direction?

Ohh yea I guess you're right I just wasn't thinking right and thought that initial displacement was zero, you're doing it right.

LeonhardEuler
Gold Member
Uhh, yeah never mind-oops!