calculus, velocity, position: a(t) = 12t - 18, v(1) = 0, x(1

[venialove]

A particle moves along the x-axis so that its acceleration at any time t > 0 is given by a(t) = 12t-18. At time t =1, the velocity of the particle is v(1) =0 and the position is x(1) = 9.

(a) write an expression for the velocity of the particle v(t)

my work for (a)
a(t) =12t-18
v(t) ?a (t) dt = ? 12t-18) dt =6t^2-18+c
0= 6(1)^2-18(1) +c
0=6-18+c
0=-12+c ?
12 = c

(b) at what values of t does the particle change direction?

my work for (b)
6t^2-18+12=0
(t^2-3t+2) =0
(t-2) (t-1) =0
t= 2,1

(c)Write an expression for the position x(t) of the particle

my work for (c)
x(t) = ? v(t) dt= ?6t^2-18t+12) dt = 2t^3- 9t-12t

(d)Find the total distance traveled by the particle from t= 3/2 to t= 6

my work for (d)
2t^3-9t^2+12t +c

 

[pka]

Re: calculus, velocity, position

\(\displaystyle \begin{gathered} x(t) = 2t^3 - 9t^2 - 12 \hfill \\ \mbox{Total distance },\,\,\int\limits_a^b {\left|{v(t)}\right|dt}\hfill\\\end{gathered}\)
Note the absolute value for total distance.