题目
Problem
6.
Two girls, Agatha and Brionie, are roller skating inside a large empty building. The girls are modelled as particles.
At time t = 0 t=0 t = 0 ( 11 i + 11 j ) m (11\mathbf{i}+11\mathbf{j})\,\mathrm{m} ( 11 i + 11 j ) m ( 7 i + 16 j ) m (7\mathbf{i}+16\mathbf{j})\,\mathrm{m} ( 7 i + 16 j ) m O O O i \mathbf{i} i j \mathbf{j} j
Agatha skates with constant velocity ( 3 i − j ) m s − 1 (3\mathbf{i}-\mathbf{j})\,\mathrm{m\,s^{-1}} ( 3 i − j ) m s − 1
Brionie skates with constant velocity ( 4 i − 2 j ) m s − 1 (4\mathbf{i}-2\mathbf{j})\,\mathrm{m\,s^{-1}} ( 4 i − 2 j ) m s − 1
(a) Find the position vector of Agatha at time t t t
(2)
At time t = 6 t=6 t = 6 P P P
(b) Show that Brionie also passes through P P P t t t
(4)
At time t t t A A A B B B
(c) Show that A B → = [ ( t − 4 ) i + ( 5 − t ) j ] m \overrightarrow{AB}=\big[(t-4)\mathbf{i}+(5-t)\mathbf{j}\big]\,\mathrm{m} A B = [ ( t − 4 ) i + ( 5 − t ) j ] m
(2)
(d) Find the distance between the two girls when they are closest together.
(4)
题目中文翻译
两个女孩 Agatha 和 Brionie 在一座宽敞的空建筑内滑旱冰。将两人视为质点。
在 t = 0 t=0 t = 0 ( 11 i + 11 j ) m (11\mathbf{i}+11\mathbf{j})\,\mathrm{m} ( 11 i + 11 j ) m ( 7 i + 16 j ) m (7\mathbf{i}+16\mathbf{j})\,\mathrm{m} ( 7 i + 16 j ) m O O O i \mathbf{i} i j \mathbf{j} j
Agatha 以恒定速度 ( 3 i − j ) m s − 1 (3\mathbf{i}-\mathbf{j})\,\mathrm{m\,s^{-1}} ( 3 i − j ) m s − 1
Brionie 以恒定速度 ( 4 i − 2 j ) m s − 1 (4\mathbf{i}-2\mathbf{j})\,\mathrm{m\,s^{-1}} ( 4 i − 2 j ) m s − 1
(a) 求 t t t
(2)
在 t = 6 t=6 t = 6 P P P
(b) 证明 Brionie 也经过 P P P t t t
(4)
在 t t t A A A B B B
(c) 证明 A B → = [ ( t − 4 ) i + ( 5 − t ) j ] m \overrightarrow{AB}=\big[(t-4)\mathbf{i}+(5-t)\mathbf{j}\big]\,\mathrm{m} A B = [ ( t − 4 ) i + ( 5 − t ) j ] m
(2)
(d) 求两人距离最近时的间距。
(4)
解答
(a)
解法一
思路
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恒速运动的位置向量等于初始位置向量加上速度向量乘时间。
答题过程
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Using r = r 0 + t v \mathbf{r}=\mathbf{r}_0+t\mathbf{v} r = r 0 + t v
r A = ( 11 i + 11 j ) + t ( 3 i − j ) m . \boxed{\mathbf{r}_A=(11\mathbf{i}+11\mathbf{j})+t(3\mathbf{i}-\mathbf{j})}\quad\mathrm{m}. r A = ( 11 i + 11 j ) + t ( 3 i − j ) m .
(b)
解法一
思路
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先算固定点 P P P
答题过程
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At t = 6 t=6 t = 6 P P P
r P = ( 11 + 18 ) i + ( 11 − 6 ) j = 29 i + 5 j . \mathbf{r}_P=(11+18)\mathbf{i}+(11-6)\mathbf{j}
=29\mathbf{i}+5\mathbf{j}. r P = ( 11 + 18 ) i + ( 11 − 6 ) j = 29 i + 5 j . Brionie’s position at time t t t
r B = ( 7 + 4 t ) i + ( 16 − 2 t ) j . \mathbf{r}_B=(7+4t)\mathbf{i}+(16-2t)\mathbf{j}. r B = ( 7 + 4 t ) i + ( 16 − 2 t ) j . To pass through P P P
7 + 4 t = 29 ⟹ t = 5.5 , 7+4t=29\quad\Longrightarrow\quad t=5.5, 7 + 4 t = 29 ⟹ t = 5.5 , 16 − 2 t = 5 ⟹ t = 5.5. 16-2t=5\quad\Longrightarrow\quad t=5.5. 16 − 2 t = 5 ⟹ t = 5.5. Both give the same time, so Brionie also passes through P P P t = 5.5 s \boxed{t=5.5\,\mathrm{s}} t = 5.5 s
(c)
解法一
思路
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从 A A A B B B B B B A A A
答题过程
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A B → = [ ( 7 + 4 t ) i + ( 16 − 2 t ) j ] − [ ( 11 + 3 t ) i + ( 11 − t ) j ] = ( 7 + 4 t − 11 − 3 t ) i + ( 16 − 2 t − 11 + t ) j = ( t − 4 ) i + ( 5 − t ) j m . \begin{align*}
\overrightarrow{AB}=&\,\big[(7+4t)\mathbf{i}+(16-2t)\mathbf{j}\big]\\[4mm]
&\,-\big[(11+3t)\mathbf{i}+(11-t)\mathbf{j}\big]\\[4mm]
=&\,(7+4t-11-3t)\mathbf{i}\\[4mm]
&\,+(16-2t-11+t)\mathbf{j}\\[4mm]
=&\,\boxed{(t-4)\mathbf{i}+(5-t)\mathbf{j}}\quad\mathrm{m}.
\end{align*} A B = = = [ ( 7 + 4 t ) i + ( 16 − 2 t ) j ] − [ ( 11 + 3 t ) i + ( 11 − t ) j ] ( 7 + 4 t − 11 − 3 t ) i + ( 16 − 2 t − 11 + t ) j ( t − 4 ) i + ( 5 − t ) j m .
(d)
解法一
思路
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距离为非负数，所以最小化距离平方即可。将二次式配方，平方项的最小值为零，并检查对应时间在允许范围内。
答题过程
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From (c), let the separation be D D D
D 2 = ( t − 4 ) 2 + ( 5 − t ) 2 = 2 t 2 − 18 t + 41 = 2 ( t − 4.5 ) 2 + 0.5. \begin{align*}
D^2=&\,(t-4)^2+(5-t)^2\\[4mm]
=&\,2t^2-18t+41\\[4mm]
=&\,2(t-4.5)^2+0.5.
\end{align*} D 2 = = = ( t − 4 ) 2 + ( 5 − t ) 2 2 t 2 − 18 t + 41 2 ( t − 4.5 ) 2 + 0.5. The square is non-negative, and it vanishes at t = 4.5 ≥ 0 t=4.5\geq0 t = 4.5 ≥ 0
D min = 0.5 m = 1 2 m ≈ 0.707 m . \boxed{D_{\min}=\sqrt{0.5}\,\mathrm{m}
=\frac1{\sqrt2}\,\mathrm{m}\approx0.707\,\mathrm{m}}. D m i n = 0.5 m = 2 1 m ≈ 0.707 m .
解法二
思路
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也可以对距离平方求导。正的二阶导数保证这个驻点是最小值，而不只是求出一个时间。
答题过程
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From (c),
D 2 = ( t − 4 ) 2 + ( 5 − t ) 2 = 2 t 2 − 18 t + 41. D^2=(t-4)^2+(5-t)^2=2t^2-18t+41. D 2 = ( t − 4 ) 2 + ( 5 − t ) 2 = 2 t 2 − 18 t + 41. Thus
d ( D 2 ) d t = 4 t − 18 = 0 ⟹ t = 4.5. \frac{\mathrm{d}(D^2)}{\mathrm{d}t}=4t-18=0
\quad\Longrightarrow\quad t=4.5. d t d ( D 2 ) = 4 t − 18 = 0 ⟹ t = 4.5. Since d 2 ( D 2 ) / d t 2 = 4 > 0 \mathrm{d}^2(D^2)/\mathrm{d}t^2=4>0 d 2 ( D 2 ) / d t 2 = 4 > 0
D min 2 = ( 0.5 ) 2 + ( 0.5 ) 2 = 0.5 , D min = 1 2 m . D_{\min}^2=(0.5)^2+(0.5)^2=0.5,
\qquad \boxed{D_{\min}=\frac1{\sqrt2}\,\mathrm{m}}. D m i n 2 = ( 0.5 ) 2 + ( 0.5 ) 2 = 0.5 , D m i n = 2 1 m .
解法三
思路
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把距离看作参数，时间方程须有实根。判别式非负给出可达到的距离下界，再检查下界确实对应一个非负时间。
答题过程
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For a separation D ≥ 0 D\geq0 D ≥ 0
2 t 2 − 18 t + ( 41 − D 2 ) = 0 2t^2-18t+(41-D^2)=0 2 t 2 − 18 t + ( 41 − D 2 ) = 0 must have a real solution. Its discriminant satisfies
( − 18 ) 2 − 4 ( 2 ) ( 41 − D 2 ) ≥ 0. (-18)^2-4(2)(41-D^2)\geq0. ( − 18 ) 2 − 4 ( 2 ) ( 41 − D 2 ) ≥ 0. Hence
324 − 328 + 8 D 2 ≥ 0 , D 2 ≥ 1 2 . 324-328+8D^2\geq0,
\qquad D^2\geq\frac12. 324 − 328 + 8 D 2 ≥ 0 , D 2 ≥ 2 1 . At equality the repeated root is t = 18 / 4 = 4.5 ≥ 0 t=18/4=4.5\geq0 t = 18/4 = 4.5 ≥ 0 D min = 1 / 2 m \boxed{D_{\min}=1/\sqrt2\,\mathrm{m}} D m i n = 1/ 2 m
