6. the position of a particle travelling along a horizontal line is s(t)=2t^3 - 3t^2 - 12t + 8, t≥0. a) find…

Explanation:

Step1: Find the velocity function

The velocity $v(t)$ is the derivative of the position - function $s(t)$. Given $s(t)=2t^{3}-3t^{2}-12t + 8$, using the power rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v(t)=s^\prime(t)=6t^{2}-6t - 12$.

Step2: Find the acceleration function

The acceleration $a(t)$ is the derivative of the velocity - function. So $a(t)=v^\prime(t)=12t-6$.

Step3: Determine when the particle is moving forwards/backwards

The particle is moving forwards when $v(t)>0$ and backwards when $v(t)<0$. First, factor $v(t)=6(t^{2}-t - 2)=6(t - 2)(t+1)$. Set $v(t)=0$, we get $t = 2$ and $t=-1$. Since $t\geq0$, we consider the intervals $[0,2)$ and $(2,\infty)$. For $t\in[0,2)$, $v(t)<0$ (the particle is moving backwards), and for $t\in(2,\infty)$, $v(t)>0$ (the particle is moving forwards).

Step4: Determine when the particle is speeding up/slowing down

The particle is speeding up when $v(t)$ and $a(t)$ have the same sign, and slowing down when they have opposite signs. Set $a(t)=0$, then $12t-6 = 0$, so $t=\frac{1}{2}$.

• For $0\leq t<\frac{1}{2}$, $a(t)<0$ and for $t>\frac{1}{2}$, $a(t)>0$.
• On the interval $[0,\frac{1}{2})$, $v(t)<0$ and $a(t)<0$, so the particle is moving backwards and speeding up.
• On the interval $(\frac{1}{2},2)$, $v(t)<0$ and $a(t)>0$, so the particle is moving backwards and slowing down.
• On the interval $(2,\infty)$, $v(t)>0$ and $a(t)>0$, so the particle is moving forwards and speeding up.

a) $v(t)=6t^{2}-6t - 12$, $a(t)=12t - 6$ b) The particle is moving forwards when $t>2$ and backwards when $0\leq t<2$. c) The particle is moving forwards and speeding up when $t>2$. d) The particle is moving backwards and speeding up when $0\leq t<\frac{1}{2}$.