A particle is moving with velocity v(t) = t^2 – 9t + 18 with distance, s measured in meters, left or right of zero, 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 zero, that is, s(0) = 1.
1:The average velocity over the interval 0 to 8 seconds
2:The time interval(s) when the particle is moving right
3:The time interval(s) when the particle is
a: going faster
b: slowing down

Question

A particle is moving with velocity v(t) = t^2 – 9t + 18 with distance, s measured in meters, left or right of zero, 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 zero, that is, s(0) = 1.
1:The average velocity over the interval 0 to 8 seconds
2:The time interval(s) when the particle is moving right
3:The time interval(s) when the particle is
a:   going faster
b:  slowing down

anonymous · Answer

The mean value theorem for integrals will help with part (1).
$$\text{average velocity}=\frac{1}{8-0}\int_0^8v(t)\,dt$$
Identically, you can solve for the position function $s(t)$, given by $s(t)=\int v(t)\,dt$ and use the initial condition. Then the mean value theorem says that the average rate of change of the position (i.e. average velocity) over the time interval would be
$$\text{average velocity}=\frac{s(8)-s(0)}{8-0}$$
You should notice that the numerators for both expressions are the same as a consequence of the fundamental theorem of calculus.

anonymous · Answer

@ can you show me the steps of how it done? i really don't know what is going on.

anonymous · Answer

What particularly are you having trouble with? Integrating? Recall the power rule:
$$\int x^n\,dx=\dfrac{1}{n+1}x^{n+1}+C$$for $n\neq-1$.
Here you have
$$\frac{1}{8}\int_0^8(t^2-9t+18)\,dt=\frac{1}{8}\left[\frac{1}{3}t^3-\frac{9}{2}t^2+18t\right]_0^8=\cdots$$
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Or, using the second approach, you solve for $s(t)$:
$$\begin{align*}s(t)&=\int v(t)\,dt\\
&=\frac{1}{3}t^3-\frac{9}{2}t^2+18t+C
\end{align*}$$When $t=0$, you have $s(0)=1$, so
$$\begin{align*}s(0)&=\frac{1}{3}0^3-\frac{9}{2}0^2+18(0)+C\\
1&=0-0+0+C\end{align*}$$So the position function is given by
$$s(t)=\frac{1}{3}t^3-\frac{9}{2}t^2+18t+1$$
Now the average velocity is the average rate of change of the position function over the time interval, so you would compute
$$\frac{s(8)-s(0)}{8-0}=\frac{\left(\dfrac{1}{3}8^3-\dfrac{9}{2}8^2+18(8)+1\right)-\left(\dfrac{1}{3}0^3-\dfrac{9}{2}0^2+18(0)+1\right)}{8}=\cdots$$

anonymous · Answer

average velocity=27/8=3.3 @SithsAndGiggles  please help me with   question 2 and 3 also!