Question

an object in rectilinear motion is modeled by the position function s(t)=3t^4 - 8t^3 - 6t^2 + 24t where s is in feet (ft) and t is in seconds (s). find the acceleration of the object when its velocity is zero. t > 0 -24 ft/s^2, 36 ft/s^2, and 72 ft/s^2 only -24 ft/s^2 and 36 ft/s^2 only 36 ft/s^2 and 72 ft/s^2 only 36 ft/s^2 only

Explanation:

Step1: Find the velocity function

The velocity function $v(t)$ is the derivative of the position function $s(t)$. Given $s(t)=3t^{4}-8t^{3}-6t^{2}+24t$, using the power - rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v(t)=s^\prime(t)=12t^{3}-24t^{2}-12t + 24$.

Step2: Set the velocity function equal to zero

Set $v(t)=0$, so $12t^{3}-24t^{2}-12t + 24 = 0$. Factor out 12: $12(t^{3}-2t^{2}-t + 2)=0$, then factor the cubic polynomial: $12[t^{2}(t - 2)-(t - 2)]=0$, which simplifies to $12(t - 2)(t^{2}-1)=0$. Further factoring $t^{2}-1=(t + 1)(t - 1)$, we get $12(t - 2)(t + 1)(t - 1)=0$. Since $t>0$, we consider $t = 1$ and $t = 2$.

Step3: Find the acceleration function

The acceleration function $a(t)$ is the derivative of the velocity function. So $a(t)=v^\prime(t)=36t^{2}-48t-12$.

Step4: Evaluate the acceleration at $t = 1$ and $t = 2$

When $t = 1$, $a(1)=36(1)^{2}-48(1)-12=36-48 - 12=-24$ ft/s².
When $t = 2$, $a(2)=36(2)^{2}-48(2)-12=144-96 - 12=36$ ft/s².

Answer:

$-24$ ft/s², $36$ ft/s²