(a) If the particle's position (measured with some unit) at time t is given by s(t), where
[tex]s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}[/tex]
then the velocity at time t, v(t), is given by the derivative of s(t),
[tex]v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}[/tex]
(b) The velocity after 3 seconds is
[tex]v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}[/tex]
(c) The particle is at rest when its velocity is zero:
[tex]\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}[/tex]
(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:
[tex]\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|<\sqrt{11} \implies 0<t<\sqrt{11}[/tex]
In interval notation, this happens for t in the interval (0, √11) or approximately (0, 3.317) s.
(e) The total distance traveled is given by the definite integral,
[tex]\displaystyle \int_0^8 |v(t)|\,\mathrm dt[/tex]
By definition of absolute value, we have
[tex]|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)<0\end{cases}[/tex]
In part (d), we've shown that v(t) > 0 when -√11 < t < √11, so we split up the integral at t = √11 as
[tex]\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt[/tex]
and by the fundamental theorem of calculus, since we know v(t) is the derivative of s(t), this reduces to
[tex]s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}[/tex]
