Question

the position of a particle moving on a horizontal line (where s is in feet to the right of a fixed reference point and t is in seconds after the start of the observation) is s(t)=t^4 - 12t + 21, t≥0. (a) find the velocity, in feet per second, at time t: v(t)=4t^3 - 12. (b) find the velocity (in ft/sec) of the particle at time t = 3. 96. (c) find all values of t for which the particle is at rest. (if there are no such values, enter none. if there are more than one value, list them separated by commas.) t = 3/√3. (d) use interval notation to indicate when the particle is moving in the positive direction. (if needed, enter inf for ∞. if the particle is never moving in the positive direction, enter none.) (√3, infinity). (e) find the total distance (in feet) traveled during the first 8 seconds.

Explanation:

Step1: Find velocity function

The position - function is $s(t)=t^{4}-12t + 21$, and the velocity function $v(t)$ is the derivative of the position function. Using the power rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v(t)=\frac{d}{dt}(t^{4}-12t + 21)=4t^{3}-12$.

Step2: Solve for part (A)

To find the velocity at $t = 3$, substitute $t = 3$ into $v(t)$.
$v(3)=4\times3^{3}-12=4\times27-12=108 - 12=96$ ft/sec.

Step3: Solve for part (B)

The particle is at rest when $v(t)=0$. So we set $4t^{3}-12 = 0$.
First, add 12 to both sides: $4t^{3}=12$.
Then divide both sides by 4: $t^{3}=3$.
Take the cube - root of both sides: $t=\sqrt[3]{3}$.

Step4: Solve for part (D)

The particle is moving in the positive direction when $v(t)>0$.
$4t^{3}-12>0$.
Add 12 to both sides: $4t^{3}>12$.
Divide both sides by 4: $t^{3}>3$.
Take the cube - root of both sides: $t>\sqrt[3]{3}$. In interval notation, the interval is $(\sqrt[3]{3},\infty)$.

Step5: Solve for part (E)

To find the total distance traveled in the first 8 seconds, we need to consider where the velocity changes sign. Since $v(t)=4t^{3}-12$ and $v(t)$ changes sign at $t = \sqrt[3]{3}$.
We know that $s(t)=t^{4}-12t + 21$.
$s(\sqrt[3]{3})=(\sqrt[3]{3})^{4}-12\sqrt[3]{3}+21=3\sqrt[3]{3}-12\sqrt[3]{3}+21=21 - 9\sqrt[3]{3}$.
$s(0)=0^{4}-12\times0 + 21=21$.
$s(8)=8^{4}-12\times8 + 21=4096-96 + 21=4021$.
The total distance $D$ is given by:
$D=\vert s(\sqrt[3]{3})-s(0)\vert+\vert s(8)-s(\sqrt[3]{3})\vert$.
$s(\sqrt[3]{3})-s(0)=21 - 9\sqrt[3]{3}-21=-9\sqrt[3]{3}$, so $\vert s(\sqrt[3]{3})-s(0)\vert=9\sqrt[3]{3}$.
$s(8)-s(\sqrt[3]{3})=4021-(21 - 9\sqrt[3]{3})=4000 + 9\sqrt[3]{3}$.
$D=9\sqrt[3]{3}+4000 + 9\sqrt[3]{3}=4000+18\sqrt[3]{3}\approx4000 + 18\times1.442=4000+25.956 = 4025.956$ ft.

Answer:

(A) 96
(B) $\sqrt[3]{3}$
(C) $\sqrt[3]{3}$
(D) $(\sqrt[3]{3},\infty)$
(E) $4000 + 18\sqrt[3]{3}$