Problem 7.11 Challenge

a) จาก \(\vec{a} = \frac{d\vec{v}}{dt}\)

îĵ

\(\vec{a} =\frac{d}{dt}((6t^{2} - 8) + 2t )\)

îĵ

\(\vec{a} = 12t + 2 m/s^{2} \;\blacksquare\)

b) จาก \(\vec{v} = \frac{d\vec{r}}{dt}\)

\(\int d\vec{r} = \int \vec{v}dt\)

îĵ

\(\int d\vec{r} = \int ((6t^{2} - 8) + 2t )dt\)

îĵ

\(\vec{r} = (2t^{3} - 8t + c_{x}) + (t^{2} + c_{y})\)

จากเวลา \(t = 0 s\) วัตถุดังกล่าวอยู่ที่ตำแหน่ง \((0 m, -4.00 m)\)

îĵîĵ

จะได้ \(0 - 4 = (2(0)^{3} - 8(0) + c_{x}) + ((0)^{2} + c_{y})\)

îĵîĵ

\(0 - 4 = c_{x}+ c_{y}\)

\(\therefore c_{x} = 0\) และ \(c_{y}= -4\)

îĵ

\(\therefore \vec{r} = (2t^{3} - 8t) + (t^{2} - 4) m \;\blacksquare\)

îĵ

c) จาก \(\vec{r}(t) = (2t^{3} - 8t) + (t^{2} - 4) m\)

îĵîĵ

จะได้ \(\vec{r}(1) = (2(1)^{3} - 8(1)) + ((1)^{2} - 4) m = -6 - 3 m\)

îĵîĵ

และ \(\vec{r}(2) = (2(2)^{3} - 8(2)) + ((2)^{2} - 4) m = 0 - 0 m\)

จาก \(\vec{v}_{av}=\frac{\Delta \vec{r}}{\Delta t}\)

\(\vec{v}_{av}=\frac{\vec{r}(2) - \vec{r}(1)}{2.00 - 1.00}\)

îĵ

\(\vec{v}_{av}= 6.00 + 3.00 m/s \;\blacksquare\)

d) วัตถุจะอยู่บนแกน y พอดี เมื่อ \(x = 0\)

จาก \(x(t) = 2t^{3} - 8t\)

\(0 = 2t^{3} - 8t\)

\(0 = t(t - 2)(t + 2)\)

\(\therefore t = \pm 2.00 s\) และ \(0 s \;\blacksquare\)