题目
Problem
7.
[In this question i \mathbf{i} i j \mathbf{j} j O O O
A ship A A A ( 2 i − 14 j ) k m h − 1 (2\mathbf{i}-14\mathbf{j})\,\mathrm{km\,h^{-1}} ( 2 i − 14 j ) km h − 1 A A A ( 8 i + 7 j ) k m (8\mathbf{i}+7\mathbf{j})\,\mathrm{km} ( 8 i + 7 j ) km
A ship B B B ( 12 i − 4 j ) k m h − 1 (12\mathbf{i}-4\mathbf{j})\,\mathrm{km\,h^{-1}} ( 12 i − 4 j ) km h − 1 B B B ( i + 2 j ) k m (\mathbf{i}+2\mathbf{j})\,\mathrm{km} ( i + 2 j ) km
(a) Show that at time t t t
B A → = [ ( 7 − 10 t ) i + ( 5 − 10 t ) j ] k m \overrightarrow{BA} = \big[(7-10t)\mathbf{i}+(5-10t)\mathbf{j}\big]\,\mathrm{km} B A = [ ( 7 − 10 t ) i + ( 5 − 10 t ) j ] km (5)
(b) Hence find the length of time for which the ships are within 2 k m 2\,\mathrm{km} 2 km
(7)
题目中文翻译
[本题中，i \mathbf{i} i j \mathbf{j} j O O O
船 A A A ( 2 i − 14 j ) k m h − 1 (2\mathbf{i}-14\mathbf{j})\,\mathrm{km\,h^{-1}} ( 2 i − 14 j ) km h − 1 A A A ( 8 i + 7 j ) k m (8\mathbf{i}+7\mathbf{j})\,\mathrm{km} ( 8 i + 7 j ) km
船 B B B ( 12 i − 4 j ) k m h − 1 (12\mathbf{i}-4\mathbf{j})\,\mathrm{km\,h^{-1}} ( 12 i − 4 j ) km h − 1 B B B ( i + 2 j ) k m (\mathbf{i}+2\mathbf{j})\,\mathrm{km} ( i + 2 j ) km
(a) 证明：下午 2 时之后 t t t
B A → = [ ( 7 − 10 t ) i + ( 5 − 10 t ) j ] k m \overrightarrow{BA} = \big[(7-10t)\mathbf{i}+(5-10t)\mathbf{j}\big]\,\mathrm{km} B A = [ ( 7 − 10 t ) i + ( 5 − 10 t ) j ] km (5)
(b) 由此求两船相距不超过 2 k m 2\,\mathrm{km} 2 km
(7)
解答
(a)
解法一
思路
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先分别写两船的位置向量。B A → \overrightarrow{BA} B A B B B A A A A A A B B B
答题过程
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At time t t t r = r 0 + t v \mathbf{r}=\mathbf{r}_0+t\mathbf{v} r = r 0 + t v
r A = ( 8 + 2 t ) i + ( 7 − 14 t ) j , \mathbf{r}_A=(8+2t)\mathbf{i}+(7-14t)\mathbf{j}, r A = ( 8 + 2 t ) i + ( 7 − 14 t ) j , r B = ( 1 + 12 t ) i + ( 2 − 4 t ) j . \mathbf{r}_B=(1+12t)\mathbf{i}+(2-4t)\mathbf{j}. r B = ( 1 + 12 t ) i + ( 2 − 4 t ) j . Hence
B A → = r A − r B = ( 8 + 2 t − 1 − 12 t ) i + ( 7 − 14 t − 2 + 4 t ) j = ( 7 − 10 t ) i + ( 5 − 10 t ) j k m . \begin{align*}
\overrightarrow{BA}=&\,\mathbf{r}_A-\mathbf{r}_B\\[4mm]
=&\,(8+2t-1-12t)\mathbf{i}\\[4mm]
&\,+ (7-14t-2+4t)\mathbf{j}\\[4mm]
=&\,\boxed{(7-10t)\mathbf{i}+(5-10t)\mathbf{j}}\quad\mathrm{km}.
\end{align*} B A = = = r A − r B ( 8 + 2 t − 1 − 12 t ) i + ( 7 − 14 t − 2 + 4 t ) j ( 7 − 10 t ) i + ( 5 − 10 t ) j km .
(b)
解法一
思路
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承接 (a) 的相对位置向量，用勾股定理求距离。先找距离为 2 2 2
答题过程
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From (a), the ships are within 2 k m 2\,\mathrm{km} 2 km
( 7 − 10 t ) 2 + ( 5 − 10 t ) 2 ≤ 4. (7-10t)^2+(5-10t)^2\leq4. ( 7 − 10 t ) 2 + ( 5 − 10 t ) 2 ≤ 4. Expanding,
49 − 140 t + 100 t 2 + 25 − 100 t + 100 t 2 ≤ 4 , 200 t 2 − 240 t + 70 ≤ 0 , 20 t 2 − 24 t + 7 ≤ 0 , ( 10 t − 7 ) ( 2 t − 1 ) ≤ 0. \begin{align*}
49-140t+100t^2+25-100t+100t^2\leq&\,4,\\[4mm]
200t^2-240t+70\leq&\,0,\\[4mm]
20t^2-24t+7\leq&\,0,\\[4mm]
(10t-7)(2t-1)\leq&\,0.
\end{align*} 49 − 140 t + 100 t 2 + 25 − 100 t + 100 t 2 ≤ 200 t 2 − 240 t + 70 ≤ 20 t 2 − 24 t + 7 ≤ ( 10 t − 7 ) ( 2 t − 1 ) ≤ 4 , 0 , 0 , 0. The boundary times are t = 1 / 2 t=1/2 t = 1/2 t = 7 / 10 t=7/10 t = 7/10
1 2 ≤ t ≤ 7 10 . \frac12\leq t\leq\frac7{10}. 2 1 ≤ t ≤ 10 7 . Its length is
7 10 − 1 2 = 1 5 h = 12 minutes . \frac7{10}-\frac12=\frac15\,\mathrm{h}
=\boxed{12\text{ minutes}}. 10 7 − 2 1 = 5 1 h = 12 minutes . 