공간에서의 운동

44.185 sec

$t=88.37$ sec

88.37 sec

$\vec v(t) = \left\langle 2,5,0\right\rangle$, $\vec a(t) = \left\langle 0,0,0\right\rangle$

$\vec v(t) = \left\langle 6t-2,-2t+1\right\rangle$, $\vec a(t) = \left\langle 6,-2 \right\rangle$

$\vec v(t) = \left\langle -\sin t, \cos t\right\rangle$, $\vec a(t) = \left\langle -\cos t, -\sin t \right\rangle$

$\vec v(t) = \left\langle 1/10, \sin t, \cos t\right\rangle$, $\vec a(t) = \left\langle 0,\cos t, -\sin t \right\rangle$

(a) Holding the crossbow at an angle of $0.013$ radians, $\approx 0.745^\circ$ will hit the target $0.4$s later. (Another solution exists, with an angle of $89^\circ$, landing $18.75$s later, but this is impractical.) (b) In the .4 seconds the arrow travels, a deer, traveling at 20mph or 29.33ft/s, can travel 11.7ft. So she needs to lead the deer by 11.7ft.

The position function is $\vec r(t) = \left\langle 220t,-16t^2+1000\right\rangle$. The $y$-component is 0 when $t=7.9$; $\vec r(7.9) = \left\langle 1739.25,0\right\rangle$, meaning the box will travel about 1740ft horizontally before it lands.

This equation stems from $\mathbf{F}=m\mathbf{a}$. In that equation, $\mathbf{a}$ is acceleration — the second derivative of position with respect to time. So, $\mathbf{v}$ is the derivative of position with respect to time.

We previously used $\mathbf{v}$ as the derivative of position with respect to the parameter we use to define our position — which was often called $t$, but was not the necessarily time. So this is a good point to keep straight.

The skateboarder starts going back down at $y_S=\frac{E}{mg}$, so we solve $3\textrm{ m}=\frac{E}{100 \mathrm{ kg} \cdot 9.8~\frac{\mathrm{m}}{\mathrm{s}^2}}$ to find $E=2940 ~\frac{\mathrm{kg}\cdot\mathrm{m}^2}{\mathrm{s}^2}=2940 J$

Remark: we needed the diameter to be greater than 3m for the skateboarder to not be going all the way around the culvert, but choosing $r=5$ leads to an answer no different from, say, $r=50$.

According to the equation in the text, the skier will become airborne when: $$|\mathbf{v}|>\sqrt{\frac{g}{\kappa}|\hat{\boldsymbol{\jmath}}\cdot\hat{\mathbf{N}}|}$$ We'll use the equation of the curve to find $\kappa$ and $\hat{\mathbf{N}}$.

Note that $g$ is given in metres per second, while the other quantities are in kilometres and hours. Converting, $9.8$ m/s$^2$ is the same as $\left(\frac{9.8\text{ m}}{1 \text{ s}^2}\right)\left( \frac{1 \text{ km}}{1000 \text{ m}}\right)\left(\frac{3600 \text{ s}}{1 \text{ hr}} \right)^2=98\cdot 6^4~ \frac{\text{km}}{\text{h}^2}=2^5\cdot 3^4\cdot 7^2~ \frac{\text{km}}{\text{h}^2}$.

$$\begin{aligned} \mathbf{r}(t)&=(\ln t,1-t)\\ \mathbf{r}'(t)&=\mathbf{v}(t)=(t^{-1}, -1)\qquad\qquad \frac{\mathrm{d}s}{\mathrm{d}t}=|\mathbf{v}(t)|=\sqrt{1+t^{-2}}\\ \mathbf{r}''(t)&=\mathbf{a}(t)=(-t^{-2},0)\\ \kappa (t)&=\frac{|\mathbf{v} \times \mathbf{a}|}{\left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^{3}}=\frac{t^{-2}}{\sqrt{1+t^{-2}}^3}=\frac{|t|}{(1+t^{2})^{3/2}}=\frac{t}{(1+t^{2})^{3/2}} \\ \text{Note $t$ is positive in the interval in question.} \\ \hat{\mathbf{T}}(t)&=\frac{\mathbf{v}(t)}{|\mathbf{v}(t)|}=\frac{1}{\sqrt{1+t^{-2}}}(t^{-1},-1)=\left( \frac{1}{\sqrt{1+t^{2}}},\frac{-t}{\sqrt{1+t^{2}}}\right)\\ \hat{\mathbf{T}}'(t)&= \left( \frac{-t}{({1+t^2})^{3/2}},\frac{-1}{({1+t^2})^{3/2}}\right)\qquad\qquad |\hat{\mathbf{T}}'(t)|=\frac{1}{t^2+1}\\ \hat{\mathbf{N}}(t)&=\frac{\hat{\mathbf{T}}'(t)}{|\hat{\mathbf{T}}'(t)|}=\left( \frac{-t}{\sqrt{1+t^2}},\frac{-1}{\sqrt{1+t^2}}\right)\\ |\hat{\mathbf{N}} \cdot \hat{\boldsymbol{\jmath}}|&=\frac{1}{\sqrt{1+t^2}} \end{aligned}$$

Now, we have all the pieces we need to find the “escape velocity" of the ground.

$$\begin{aligned} |\mathbf{v}|&=\sqrt{\frac{g}{\kappa}|\hat{\mathbf{N}} \cdot \hat{\boldsymbol{\jmath}}|}=\sqrt{\frac{g\cdot (1+t^{2})^{3/2}}{t(1+t^2)^{1/2}}}=\sqrt{\frac{g(1+t^2)}{t}} \end{aligned}$$

Since the skier can take off anywhere on the hill, we just need their velocity to be larger than the smallest value of $\sqrt{\frac{g(1+t^2)}{t}}$ when $1/e \le t \le e$. To find that minimum, we find the location of the minimum of the simpler function $g(t)=\frac{1+t^2}{t}$. Using first-semester calculus, we find it to occur when $t=1$. So, the minimum value of $\sqrt{\frac{g(1+t^2)}{t}}$ (that is, smallest speed to achieve lift-off) occurs at $t=1$. We therefore need a minimum speed greater than:

$$\begin{aligned} \sqrt{\frac{g(1+t^2)}{|t|}}\Bigg|_{t=1}&=\sqrt{2g}=\sqrt{2^6\cdot3^4\cdot7^2}=2^3\cdot 3^2\cdot 7 = 504\text{ kph} \end{aligned}$$

(It seems unlikely that one could reach this speed on skis. The skier is probably earth-bound until they find a curvier hill.)