Question

A particle moves along the x-axis with the given acceleration function a(t) = 8t - 6, initial position s(0) = 2, and initial velocity v(0) = 0. Find the position function. Thanks(:

Answer 1

First you want to integrate the acceleration function:
\[\int\limits_{}^{}8t-6dt = 4t^2-6t+c\]
this is the velocity, and since our initial conditions is v(0) - 0, we get that the constant is 0 so now we have:
\[v(t) = 4t^2-6t\]

Answer 2

Now we have to do the same process again. We need to integrate the velocity function to get the position function:
\[\int\limits_{}^{}4t^2-6tdt = \frac{4}{3}t^3-3t^2+c\]
and because the initial conditions s(0) = 2, c must be 2, so the position function is:
\[s(t) = \frac{4}{3}t^3-3t^2+2\]

Answer 3

Ohhh I get it now! I looked on paul's open study notes and it cleared my confusion to your answer! (:

Answer 4

looks like joes got it wrapped up :) but lets verify

"A particle moves along the x-axis with the given acceleration function a(t) = 8t - 6, initial position s(0) = 2, and initial velocity v(0) = 0. Find the position function. Thanks (:"

integrate: 8t - 6 to get v(t), thats just the convention...

v(t) = 8t^2/2 -6t +C ; and we are told that v(0) = 0

0 = 4(0)^2 - 6(0) + C

0 = C, which means

v(t) = 4t^2 -6t, integrate again

integrate: 4t^2 -6t to get s(t), which is the position function

s(t) = 4t^3/3 -6t^2/2 + C ; now it tells us that s(0) = 2

2 = 4(0)^3/3 -3(0)^2/2 + C

2 = C , lets fill in the pieces now..

s(t) = \(\cfrac{4}{3}t^3-3t^2+2\)

and Joe is verified :)

Answer 5

Good answer Joe, you deserved the medal