Question

a particle is moving along a straight line and its position with respect to a reference point is s = 2t^3 - 7t^2 - 8t + 14 (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. 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 $s = 2t^{3}-7t^{2}-8t + 14$, by the power - rule of differentiation $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v(t)=\frac{ds}{dt}=6t^{2}-14t - 8$.

Step2: Recall acceleration - velocity relationship

Acceleration $a(t)$ is the derivative of velocity $v(t)$. Differentiating $v(t)=6t^{2}-14t - 8$ using the power - rule, we get $a(t)=\frac{dv}{dt}=12t-14$.

Step3: Find acceleration at $t = 1$

Substitute $t = 1$ into the acceleration formula $a(t)$. So $a(1)=12\times1-14=-2$ m/s².

Step4: Find when velocity is 0

Set $v(t)=6t^{2}-14t - 8 = 0$. Divide through by 2: $3t^{2}-7t - 4=0$. Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $ax^{2}+bx + c = 0$, here $a = 3$, $b=-7$, $c=-4$. Then $t=\frac{7\pm\sqrt{(-7)^{2}-4\times3\times(-4)}}{2\times3}=\frac{7\pm\sqrt{49 + 48}}{6}=\frac{7\pm\sqrt{97}}{6}$. We have $t=\frac{7+\sqrt{97}}{6}\approx\frac{7 + 9.85}{6}\approx2.81$ and $t=\frac{7-\sqrt{97}}{6}\approx\frac{7 - 9.85}{6}\approx - 0.48$. Since $t\geq0$, we take $t=\frac{7+\sqrt{97}}{6}\approx2.81$. Substitute this value of $t$ into the acceleration formula $a(t)=12t-14$. So $a(\frac{7 + \sqrt{97}}{6})=12\times\frac{7+\sqrt{97}}{6}-14=14 + 2\sqrt{97}-14=2\sqrt{97}\approx19.70$ m/s².

Answer:

a. $v(t)=6t^{2}-14t - 8$, $a(t)=12t-14$
b. $a(1)=-2$ m/s²
c. $a\approx19.70$ m/s²