[tex]x(t)=7.00\,\mathrm m-\left(9.00\dfrac{\rm m}{\rm s}\right)t+\left(3\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]
a. The particle has velocity at time [tex]t[/tex],
[tex]\dfrac{\mathrm dx(t)}{\mathrm dt}=-9.00\dfrac{\rm m}{\rm s}+\left(6\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]
so that after [tex]t=1\,\mathrm s[/tex] it will have velocity [tex]\boxed{-3.00\dfrac{\rm m}{\rm s}}[/tex].
b. The sign of the velocity is negative, so it's moving in the negative [tex]x[/tex] direction.
c. Its speed is 3.00 m/s.
d. The particle's velocity changes according to
[tex]\dfrac{\mathrm d^2x(t)}{\mathrm dt^2}=6\dfrac{\rm m}{\mathrm s^2}[/tex]
which is positive and indicates the velocity/speed of the particle is increasing.
e. Yes. The velocity is increasing at a constant rate. Solving for [tex]\dfrac{\mathrm dx(t)}{\mathrm dt}=0[/tex] is trivial; this happens when [tex]\boxed{t=1.50\,\mathrm s}[/tex].
f. No, the velocity is positive for all [tex]t[/tex] beyond 1.50 s.
