(d) The particle moves in the positive direction when its velocity has a positive sign. You know the particle is at rest when [tex]t=0[/tex] and [tex]t=3[/tex], and because the velocity function is continuous, you need only check the sign of [tex]v(t)[/tex] for values on the intervals (0, 3) and (3, 6).
We have, for instance [tex]v(1)\approx-0.91<0[/tex] and [tex]v(4)\approx0.91>0[/tex], which means the particle is moving the positive direction for [tex]3<t<6[/tex], or the interval (3, 6).
(e) The total distance traveled is obtained by integrating the absolute value of the velocity function over the given interval:
[tex]\displaystyle\int_0^6|v(t)|\,\mathrm dt=\int_0^3-v(t)\,\mathrm dt+\int_3^6v(t)\,\mathrm dt[/tex]
which follows from the definition of absolute value. In particular, if [tex]x[/tex] is negative, then [tex]|x|=-x[/tex].
The total distance traveled is then 4 ft.
(g) Acceleration is the rate of change of velocity, so [tex]a(t)[/tex] is the derivative of [tex]v(t)[/tex]:
[tex]a(t)=v'(t)=-\dfrac{\pi^2}9\cos\left(\dfrac{\pi t}3\right)[/tex]
Compute the acceleration at [tex]t=4[/tex] seconds:
[tex]a(t)=\dfrac{\pi^2}{18}\dfrac{\rm ft}{\mathrm s^2}[/tex]
(In case you need to know, for part (i), the particle is speeding up when the acceleration is positive. So this is done the same way as part (d).)
