Question

PLEASE HELP ASAP: A particle is moving with velocity v(t) = t2 – 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.
The average velocity over the interval 0 to 8 seconds

The instantaneous velocity and speed at time 5 secs

The time interval(s) when the particle is moving right

The time interval(s) when the particle is
going faster
slowing down

Find the total distance the particle has traveled between 0 and 8 seconds

Answer 1

1) $\frac{v(8)-v(0)}{8 - 0} = \frac{10-18}{8} = -1\frac{m}{s}$.
2) $v(5) = 5^2-9*5+18 = 25-45+18 = -2\frac{m}{s}$.
3) The particle is moving right when the velocity function is positive: $0\ \textless \ t\ \textless \ 3$ or $6\ \textless \ t\ \textless \ 8$.
4) When $0\ \textless \ t\ \textless \ \frac{9}{2}$ the particle is slowing down because the acceleration is close to zero $\Rightarrow$ the particle is speeding up when acceleration is increasing away from zero: $\frac{9}{2}\ \textless \ t\ \textless \ 8$.
5) $(\frac{1}{8})* \int\limits^8_0 {t^2-9t+18dt}=\frac{1}{8}*(\frac{t^3}{3}-(\frac{9}{2}*t^2+18t)_{0}^{8}= \\=(\frac{1}{8})*(\frac{8^3}{3}-(\frac{9}{2})*8^2+18*8)=\frac{8^2}{3}-(\frac{9}{2})*8+18=3\frac{1}{3} \frac{m}{s}$.