Given that the velocity at any time t is
[tex]v(t)=t^2-9t+18[/tex]
Also, the time interval is from t = 0 to t = 8 seconds
The position at time t = 0 s is s(0) = 1 m towards right of zero.
The initial time is t = 0 s, so the initial velocity will be
[tex]\begin{gathered} v_i(t=0)=0^2-9\times0+18\text{ } \\ v_i(0)\text{ = 18 m/s} \end{gathered}[/tex]
The final time is t = 8 s, so the final velocity will be
[tex]\begin{gathered} v_f(t=8)=8^2-9\times8+18 \\ v_f(8)\text{ = 64-72+18} \\ =\text{ 10 m/s} \end{gathered}[/tex]
The average velocity will be
[tex]\begin{gathered} v_{av}=\frac{v_i+v_f}{2} \\ =\frac{18+10}{2} \\ =14\text{ m/s} \end{gathered}[/tex]
Thus, the average velocity is 14 m/s.
Part II:
The instantaneous velocity at time t =5 s will be
[tex]\begin{gathered} v(t=5)=5^2-9\times5+18 \\ =25-45+18 \\ =-2\text{ m/s} \end{gathered}[/tex]
The instantaneous speed is the magnitude of instantaneous velocity.
Thus, the instantaneous speed will be 2 m/s.
Part III:
The particle will move towards the right when v(t) > 0
The time intervals will be
[tex]\begin{gathered} t^2-9t+18>0 \\ t^2-6t-3t+18>0 \\ t(t-6)-3(t-6)>0 \\ (t-6)(t-3)>0 \\ t-6>0\text{ or t>6} \\ t-3>0\text{ ot t>3} \end{gathered}[/tex]
Thus, time intervals are t > 3 and t > 6 when the particle is moving towards the right.
Part IV :
The particle will move faster if the acceleration, a(t) > 0
The particle will slow down if the acceleration, a(t) < 0
So, first, we need to find the acceleration, it can be calculated as
[tex]\begin{gathered} a(t)=\text{ }\frac{d(v(t))}{dt} \\ =\frac{d(t^2-9t+18)}{dt} \\ =2t-9 \end{gathered}[/tex]
For the particle moving faster,
[tex]\begin{gathered} a(t)>0 \\ 2t-9>0 \\ 2t-9+9>9+0 \\ 2t>9 \\ \frac{2t}{2}>\frac{9}{2} \\ t>\frac{9}{2} \\ t>4.5\text{ s} \end{gathered}[/tex]
For particle slowing down,
[tex]\begin{gathered} a(t)<0 \\ 2t-9<0 \\ 2t-9+9<9+0_{} \\ 2t<9 \\ \frac{2t}{2}<\frac{9}{2} \\ t<4.5\text{ s} \end{gathered}[/tex]
The total distance can be calculated as
[tex]\begin{gathered} s(t)=\int ^8_0v(t)dt \\ =\text{ }\int ^8_0(t^2-9t+18)\mathrm{d}t \\ =\lbrack\frac{t^3}{3}\rbrack^8_0-9\lbrack\frac{t^2}{2}\rbrack^8_0+18\lbrack t^{}\rbrack^8_0 \\ =\frac{1}{3}\lbrack512-0\rbrack-9\lbrack64-0\rbrack+18\lbrack8-0\rbrack \\ =\text{ 170.67-576+144} \\ =-261.33\text{ m} \end{gathered}[/tex]
Here, the negative symbol indicates it is towards the left from zero.
