บทที่ 3 · แรงและกฎการเคลื่อนที่
Problem 3.57 Challenge
หัวข้อ: แรงขึ้นกับเวลา (Time Dependent Forces)
\(\therefore \vec{r}(t) = (\frac{t^{3}}{6}+ 5.0) -\frac{4.9t^{2}}{2}m \;\blacksquare\)
ดูวิธีทำ
a) จาก \(\vec{F} = m\vec{a}\)
\(\vec{a} =\frac{\vec{F}(t)}{m}\)
îĵ
\(\vec{a}(t) = t - 4.9 m/s^{2} \;\blacksquare\)
b) จาก \(\vec{v}(t) = \int \vec{a}(t)dt\)
îĵ
\(\vec{v}(t) = \int (t - 4.9 )dt\)
îĵ
\(\vec{v}(t) = (\frac{t^{2}}{2}+ c_{x}) + (c_{y}- 4.9t)\)
จาก \(\vec{v}(0) = 0 m/s\)
îĵîĵ
\(c_{x}+ c_{y}= 0 + 0\to c_{x} = 0\) และ \(c_{y}= 0\)
îĵ
\(\therefore \vec{v}(t) =\frac{t^{2}}{2}- 4.9t m/s \;\blacksquare\)
c) จาก \(\vec{r}(t) = \int \vec{v}(t)dt\)
îĵ
\(\vec{r}(t) = \int (\frac{t^{2}}{2}- 4.9t )dt\)
îĵ
\(\vec{r}(t) = (\frac{t^{3}}{6}+ c_{x}) + (c_{y}-\frac{4.9t^{2}}{2})\)
î
จาก \(\vec{r}(0) = 5.0 m\)
îĵîĵ
\(c_{x}+ c_{y}= 5.0 + 0\to c_{x} = 5.0\) และ \(c_{y}= 0\)
îĵ
\(\therefore \vec{r}(t) = (\frac{t^{3}}{6}+ 5.0) -\frac{4.9t^{2}}{2}m \;\blacksquare\)