Question

a particle moves according to the law of motion s = t^3 - 8t^2 + 2t, t ≥ 0, where t is measured in seconds and s in feet. a.) find the velocity at time t. answer: b.) what is the velocity after 3 seconds? answer: c.) when is the particle at rest? enter your answer as a comma separated list. enter none if the particle is never at rest. answer: d.) when is the particle moving in the positive direction? when 0 ≤ t < and t >

Explanation:

Step1: Recall velocity - displacement relation

Velocity $v(t)$ is the derivative of displacement $s(t)$. Given $s(t)=t^{3}-8t^{2}+2t$, by the power - rule $\frac{d}{dt}(t^{n}) = nt^{n - 1}$, we have $v(t)=\frac{d}{dt}(t^{3}-8t^{2}+2t)$.
$v(t)=3t^{2}-16t + 2$

Step2: Find velocity at $t = 3$

Substitute $t = 3$ into $v(t)$.
$v(3)=3\times3^{2}-16\times3 + 2=3\times9-48 + 2=27-48 + 2=-19$

Step3: Find when particle is at rest

The particle is at rest when $v(t)=0$. So we solve the quadratic equation $3t^{2}-16t + 2=0$. Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $ax^{2}+bx + c = 0$. Here $a = 3$, $b=-16$, $c = 2$.
$t=\frac{16\pm\sqrt{(-16)^{2}-4\times3\times2}}{2\times3}=\frac{16\pm\sqrt{256 - 24}}{6}=\frac{16\pm\sqrt{232}}{6}=\frac{16\pm2\sqrt{58}}{6}=\frac{8\pm\sqrt{58}}{3}$

Step4: Find when particle moves in positive direction

The particle moves in the positive direction when $v(t)>0$. The roots of $v(t)=3t^{2}-16t + 2$ are $t_1=\frac{8-\sqrt{58}}{3}$ and $t_2=\frac{8 + \sqrt{58}}{3}$. Since the coefficient of $t^{2}$ in $v(t)$ is positive ($a = 3>0$), the quadratic function $y = v(t)$ is a parabola opening upwards. So $v(t)>0$ when $0\leq t<\frac{8 - \sqrt{58}}{3}$ and $t>\frac{8+\sqrt{58}}{3}$

Answer:

a. $3t^{2}-16t + 2$
b. $-19$
c. $\frac{8 - \sqrt{58}}{3},\frac{8+\sqrt{58}}{3}$
d. $\frac{8 - \sqrt{58}}{3},\frac{8+\sqrt{58}}{3}$