Question

Please help me, I am following along diligently. V(t) = t^2 -9t+18, with distance, s measured in meters, left or right of 0, and t measured in seconds, with t between 0 and 8 seconds inclusive. The position at time t=0 sec is 1 meter right of 0, that is s(0)= 1Part I: Average velocity over the interval 0 to 8 secondsPart II: The instantaneous velocity and speed at time 5 secsPart III: The time intervals when the particle is moving rightPart IV: The time intervals when the particle is going faster, and slowing downPart V: Total distance the particle has traveled between 0 and 8 seconds

Answer 1

Given that the velocity at any time t is

`v(t)=t^2-9t+18`

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

`\begin{gathered} v_i(t=0)=0^2-9*0+18\text{ } \\ v_i(0)\text{ = 18 m/s} \end{gathered}`

The final time is t = 8 s, so the final velocity will be

`\begin{gathered} v_f(t=8)=8^2-9*8+18 \\ v_f(8)\text{ = 64-72+18} \\ =\text{ 10 m/s} \end{gathered}`

The average velocity will be

`\begin{gathered} v_(av)=(v_i+v_f)/(2) \\ =(18+10)/(2) \\ =14\text{ m/s} \end{gathered}`

Thus, the average velocity is 14 m/s.

Part II:

The instantaneous velocity at time t =5 s will be

`\begin{gathered} v(t=5)=5^2-9*5+18 \\ =25-45+18 \\ =-2\text{ m/s} \end{gathered}`

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

`\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}`

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

`\begin{gathered} a(t)=\text{ }(d(v(t)))/(dt) \\ =(d(t^2-9t+18))/(dt) \\ =2t-9 \end{gathered}`

For the particle moving faster,

`\begin{gathered} a(t)>0 \\ 2t-9>0 \\ 2t-9+9>9+0 \\ 2t>9 \\ (2t)/(2)>(9)/(2) \\ t>(9)/(2) \\ t>4.5\text{ s} \end{gathered}`

For particle slowing down,

`\begin{gathered} a(t)<0 \\ 2t-9<0 \\ 2t-9+9<9+0_{} \\ 2t<9 \\ (2t)/(2)<(9)/(2) \\ t<4.5\text{ s} \end{gathered}`

The total distance can be calculated as

`\begin{gathered} s(t)=\int ^8_0v(t)dt \\ =\text{ }\int ^8_0(t^2-9t+18)\mathrm{d}t \\ =\lbrack(t^3)/(3)\rbrack^8_0-9\lbrack(t^2)/(2)\rbrack^8_0+18\lbrack t^{}\rbrack^8_0 \\ =(1)/(3)\lbrack512-0\rbrack-9\lbrack64-0\rbrack+18\lbrack8-0\rbrack \\ =\text{ 170.67-576+144} \\ =-261.33\text{ m} \end{gathered}`

Here, the negative symbol indicates it is towards the left from zero.