a) The position of the particle along the straight-line path is given by
[tex]s=(t^3-6t^2-15 t+7)ft[/tex]
where t is the time.
If we substitute t=11 s inside the equation, we find the total distance covered during this time, in feet:
[tex]s=(11)^3-6(11)^2-15(11)+7=597 ft[/tex]
which converted into meters is s=182 m.
b) The average velocity of the particle at t=11 s is equal to the total distance travelled, s=597 ft, divided by the time taken, t=11 s:
[tex]v= \frac{s}{t}= \frac{597 ft}{11s} =54.3 ft/s [/tex]
We can also calculate it in meters/second:
[tex]v= \frac{s}{t}= \frac{182 m}{11 s}=16.5 m/s [/tex]
