Question

a particle moves according to the law of motion s = t^3 - 8t^2 + 4t, t ≥ 0, where t is measured in seconds and s in feet. a.) find the velocity at time t. b.) what is the velocity after 3 seconds? c.) when is the particle at rest? enter your answer as a comma separated list. enter none if the particle is never at rest. d.) when is the particle moving in the positive direction? when 0 ≤ t < and t >

Explanation:

Step1: Recall the velocity - position relationship

The velocity $v(t)$ of a particle is the derivative of the position function $s(t)$. Given $s(t)=t^{3}-8t^{2}+4t$, by the power - rule of differentiation $\frac{d}{dt}(t^{n}) = nt^{n - 1}$, we have $v(t)=\frac{ds}{dt}=3t^{2}-16t + 4$.

Step2: Find the velocity at $t = 3$

Substitute $t = 3$ into $v(t)$.
$v(3)=3(3)^{2}-16(3)+4$.
$v(3)=3\times9-48 + 4$.
$v(3)=27-48 + 4=-17$ feet per second.

Step3: Find when the particle is at rest

The particle is at rest when $v(t)=0$. So we set $3t^{2}-16t + 4 = 0$.
Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a = 3$, $b=-16$, and $c = 4$.
$t=\frac{16\pm\sqrt{(-16)^{2}-4\times3\times4}}{2\times3}=\frac{16\pm\sqrt{256 - 48}}{6}=\frac{16\pm\sqrt{208}}{6}=\frac{16\pm4\sqrt{13}}{6}=\frac{8\pm2\sqrt{13}}{3}$.
$t_1=\frac{8 + 2\sqrt{13}}{3}\approx\frac{8+2\times3.606}{3}=\frac{8 + 7.212}{3}\approx5.07$ and $t_2=\frac{8 - 2\sqrt{13}}{3}\approx\frac{8-7.212}{3}\approx0.26$.

Step4: Determine when the particle is moving in the positive direction

The particle is moving in the positive direction when $v(t)>0$.
We know $v(t)=3t^{2}-16t + 4=(3t - 2)(t - 2)$.
The roots of $v(t)$ are $t=\frac{2}{3}$ and $t = 2$.
We can test the intervals:
For $0\leq t<\frac{2}{3}$, let $t=\frac{1}{3}$, then $v(\frac{1}{3})=3\times(\frac{1}{3})^{2}-16\times\frac{1}{3}+4=\frac{1}{3}-\frac{16}{3}+4=\frac{1 - 16 + 12}{3}=-1<0$.
For $\frac{2}{3}For $t>2$, let $t = 3$, then $v(3)=-17<0$.
The particle is moving in the positive direction when $02$.

Answer:

a. $v(t)=3t^{2}-16t + 4$
b. $-17$ feet per second
c. $\frac{8 + 2\sqrt{13}}{3},\frac{8 - 2\sqrt{13}}{3}$
d. $02$