Math2

速度是位移對時間的微分

\(\bar{v} = \frac{x_2-x_1}{t_2-t_1}=\frac{\Delta x}{\Delta t}　(xt圖的斜率)\)
\(when \underset{\Delta t\rightarrow0}\lim v = \frac{dx}{dt}\)
\begin{cases} x'(t)=v(t)\\ \int{v(t)dt} = x(t) 　(vt圖的面積) \end{cases}
換句話說：位移是速度對時間的積分

\(\bar{v} = \frac{x_2-x_1}{t_2-t_1}=\frac{\Delta x}{\Delta t}\)

瞬時速度：\(v=\underset{\Delta t\rightarrow0}\lim{\frac{\Delta x}{\Delta t}}\)

加速度是速度對時間的微分

\(\bar{a} = \frac{v_2-v_1}{t_2-t_1}=\frac{\Delta v}{\Delta t}　(vt圖的斜率)\)
\(when \underset{\Delta t\rightarrow0}\lim a = \frac{dv}{dt}\)
\begin{cases} v'(t)=a(t)\\ \int{a(t)dt} = v(t) 　 \end{cases}
換句話說：速度是加速度對時間的積分

總結

\begin{aligned} &\begin{cases} x'(t)=v(t)\\ \int{v(t)dt} = x(t) \end{cases}\\\\ &\begin{cases} v'(t)=a(t)\\ \int{a(t)dt} = v(t) 　 \end{cases} \end{aligned}

已知：\(H, T_{max}, T_{total}\)
思路：從 "已知" 推出 "未知"
可以分為兩個部分：

1. 石頭向上移動到最高點
2. 石頭從向上到向下的整個運動過程

要怎麼將1與2取得關聯呢？
之間的關係是什麼？\(3T_{max}=T_{total}\)

1. 石頭向上移動到最高點，得到時間與速度的關係：
   \(V=V_0+at\)
   得到\(V=V_0+gT_{max}\)
   \(T_{max}=\frac{V_0}{g}\)
   有了\(T_{max}就能從3T_{max}=T_{total}\)得到整體時間的關係

2. 石頭從向上到向下的整個運動過程：
   

有什麼資訊？

從1.得到時間的關係：
(1) \(T_{max}=\frac{V_0}{g}\) 　
(2) \(3T_{max}=T_{total}　\)

2. 石頭從向上到向下的整個運動過程：(向上為正，向下為負。)
   \(S=V_0t+\frac{1}{2}at^2\)　得到\(-H=V_0T_{total}-\frac{1}{2}gT_{total}^2\)

\begin{aligned} (2)\because 3T_{max}&=T_{total}\\ \therefore -H&=V_03T_{max}-\frac{1}{2}g(3T_{max})^2\\ (1)\because T_{max}&=\frac{V_0}{g}\\ \therefore -H&=\frac{3V_0^2}{g}-\frac{9V_0^2}{2g}=\frac{-3V_0^2}{2g}\\ V_0^2&=\frac{2Hg}{3} \Rightarrow V_0=\sqrt{\frac{2Hg}{3}}\\ \end{aligned}

微積分應用

Part A
Find its position at the first instant when the car has zero velocity.

\(x'(t)=v(t)=(9.60m/s^2)t-(0.600m/s^6)t^5\)
\(if:v(t)=0,t=0\) (the first instant)
x(0)=2.17m

Part B
Find its acceleration at the first instant when the car has zero velocity.

\(x'(t)=v(t)=(9.60m/s^2)t-(0.600m/s^6)t^5\)
\(v'(t)=a(t)=(9.60m/s^2)-(3.000m/s^6)t^4\)

\(if:v(t)=0,t=0\) (the first instant)
\(a(0)=9.60m/s^2\)

Part C
Find its position at the second instant when the car has zero velocity.

\(x'(t)=v(t)=(9.60m/s^2)t-(0.600m/s^6)t^5\)
\(v(t)=(9.60m/s^2)t-(0.600tm/s^6)t^5\)

\(if:v(t)=0,(9.60m/s^2)t=(0.600tm/s^6)t^5\)
\(t^4=\frac{9.60m/s^2}{0.600tm/s^6}\)
t=2 (the second instant)
x(2)=14.97m

Part D
Find its acceleration at the second instant when the car has zero velocity.

\(v'(t)=a(t)=(9.60m/s^2)-(3.000m/s^6)t^4\)
t=2 (the second instant)
\(a(2)=38.4m/s^2\)

三角函數

\(\vec{a}\cdot\vec{b}=\vert a\vert\vert b\vert\cos\theta\)
\(\vert\vec{a}\times\vec{b}\vert=\vert a\vert\vert b\vert\sin\theta\)

\begin{aligned} \vec{a}\cdot\vec{b}&=6\vert\vec{a}\times\vec{b}\vert\\ \vert a\vert\vert b\vert\cos\theta&=6\vert a\vert\vert b\vert\sin\theta\\ cos\theta&=6sin\theta\\ \frac{1}{6}&=\frac{sin\theta}{cos\theta}=tan\theta\\ arctan(\frac{1}{6})&\approx 9.4623° \end{aligned}

Part C
drives 3.40 km north, then 2.35 km west, and then 1.30 km south.

\(90 -arcsin(\frac{2.10}{\sqrt{2.35^2+2.1^2}})\approx 48.215°\)

後記

\begin{cases} v=v_0+at\\ s=v_0t+\frac{1}{2}at^2\\ v^2=v_0^2+2as　 \end{cases}

微積分應用