Question

a particle is moving along a straight line and its position with respect to a reference point is s = 2t^3 - 9t^2 - 6t + 13 (where s is in meters and t is in seconds, and assume that t is non - negative). (round all decimal answers to 2 decimal places.) a. find the velocity and acceleration as functions of t. v(t)= a(t)= b. find the acceleration after 1 second. c. find the acceleration at the instant when the velocity is 0.

Explanation:

Step1: Recall velocity - position relationship

Velocity $v(t)$ is the derivative of position $s(t)$. Given $s(t)=2t^{3}-9t^{2}-6t + 13$, using the power - rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v(t)=\frac{d}{dt}(2t^{3}-9t^{2}-6t + 13)=6t^{2}-18t - 6$.

Step2: Recall acceleration - velocity relationship

Acceleration $a(t)$ is the derivative of velocity $v(t)$. Differentiating $v(t)=6t^{2}-18t - 6$ with the power - rule, we get $a(t)=\frac{d}{dt}(6t^{2}-18t - 6)=12t-18$.

Step3: Find acceleration at $t = 1$

Substitute $t = 1$ into the acceleration function $a(t)$. So, $a(1)=12\times1-18=-6.00$.

Step4: Find when velocity is 0

Set $v(t)=0$, so $6t^{2}-18t - 6 = 0$. Divide through by 6 to get $t^{2}-3t - 1=0$. Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $at^{2}+bt + c = 0$ (here $a = 1$, $b=-3$, $c=-1$), we have $t=\frac{3\pm\sqrt{(-3)^{2}-4\times1\times(-1)}}{2\times1}=\frac{3\pm\sqrt{9 + 4}}{2}=\frac{3\pm\sqrt{13}}{2}$. Since $t\geq0$, $t=\frac{3+\sqrt{13}}{2}\approx3.30$ (we take the positive root as $t\geq0$).

Step5: Find acceleration when velocity is 0

Substitute $t=\frac{3+\sqrt{13}}{2}$ into the acceleration function $a(t)$. $a(\frac{3+\sqrt{13}}{2})=12\times\frac{3+\sqrt{13}}{2}-18=18 + 6\sqrt{13}-18=6\sqrt{13}\approx21.63$.

Answer:

a. $v(t)=6t^{2}-18t - 6$
$a(t)=12t-18$
b. $-6.00$
c. $21.63$