题目
Problem
8.
[In this question, i \mathbf{i} i j \mathbf{j} j O O O
Two ships, A A A B B B
The velocity of A A A ( 3 i + 12 j ) k m h − 1 (3\mathbf{i}+12\mathbf{j})\,\mathrm{km\,h^{-1}} ( 3 i + 12 j ) km h − 1 B B B ( p i + q j ) k m h − 1 (p\mathbf{i}+q\mathbf{j})\,\mathrm{km\,h^{-1}} ( p i + q j ) km h − 1
(a) Find the speed of A A A
(2)
The ships are modelled as particles.
At 12 noon, A A A ( − 9 i + 6 j ) k m (-9\mathbf{i}+6\mathbf{j})\,\mathrm{km} ( − 9 i + 6 j ) km B B B ( 16 i + 6 j ) k m (16\mathbf{i}+6\mathbf{j})\,\mathrm{km} ( 16 i + 6 j ) km
At time t t t
A B → = [ ( 25 − 12 t ) i − 9 t j ] k m \overrightarrow{AB}=\big[(25-12t)\mathbf{i}-9t\mathbf{j}\big]\,\mathrm{km} A B = [ ( 25 − 12 t ) i − 9 t j ] km (b) Find the value of p p p q q q
(7)
(c) Find the bearing of A A A B B B 15 k m 15\,\mathrm{km} 15 km
(7)
题目中文翻译
[本题中，i \mathbf{i} i j \mathbf{j} j O O O
两艘船 A A A B B B
A A A ( 3 i + 12 j ) k m h − 1 (3\mathbf{i}+12\mathbf{j})\,\mathrm{km\,h^{-1}} ( 3 i + 12 j ) km h − 1 B B B ( p i + q j ) k m h − 1 (p\mathbf{i}+q\mathbf{j})\,\mathrm{km\,h^{-1}} ( p i + q j ) km h − 1
(a) 求 A A A
(2)
将两艘船视为质点。
正午 12 时，A A A ( − 9 i + 6 j ) k m (-9\mathbf{i}+6\mathbf{j})\,\mathrm{km} ( − 9 i + 6 j ) km B B B ( 16 i + 6 j ) k m (16\mathbf{i}+6\mathbf{j})\,\mathrm{km} ( 16 i + 6 j ) km
正午 12 时之后 t t t
A B → = [ ( 25 − 12 t ) i − 9 t j ] k m \overrightarrow{AB}=\big[(25-12t)\mathbf{i}-9t\mathbf{j}\big]\,\mathrm{km} A B = [ ( 25 − 12 t ) i − 9 t j ] km (b) 求 p p p q q q
(7)
(c) 求两船相距 15 k m 15\,\mathrm{km} 15 km A A A B B B
(7)
解答
(a)
解法一
思路
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速率是速度向量的模，取两个互相垂直分量的平方和再开方。
答题过程
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The speed of A A A
∣ v A ∣ = 3 2 + 12 2 = 153 = 3 17 k m h − 1 . |\mathbf v_A|=\sqrt{3^2+12^2}=\sqrt{153}=\boxed{3\sqrt{17}\ \mathrm{km\,h^{-1}}}. ∣ v A ∣ = 3 2 + 1 2 2 = 153 = 3 17 km h − 1 .
(b)
解法一
思路
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分别写两船位置再相减，得到相对位置中时间项的系数。与题给表达式比较两个分量，求出 p p p q q q
答题过程
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At t t t
a = ( − 9 + 3 t ) i + ( 6 + 12 t ) j , b = ( 16 + p t ) i + ( 6 + q t ) j . \begin{align*}
\mathbf a=&\,(-9+3t)\mathbf i+(6+12t)\mathbf j,\\
\mathbf b=&\,(16+pt)\mathbf i+(6+qt)\mathbf j.
\end{align*} a = b = ( − 9 + 3 t ) i + ( 6 + 12 t ) j , ( 16 + pt ) i + ( 6 + q t ) j . Therefore
A B → = b − a , = [ 25 + ( p − 3 ) t ] i + ( q − 12 ) t j . \begin{align*}
\overrightarrow{AB}=&\,\mathbf b-\mathbf a,\\
=&\,\bigl[25+(p-3)t\bigr]\mathbf i+(q-12)t\mathbf j.
\end{align*} A B = = b − a , [ 25 + ( p − 3 ) t ] i + ( q − 12 ) t j . Comparing with ( 25 − 12 t ) i − 9 t j (25-12t)\mathbf i-9t\mathbf j ( 25 − 12 t ) i − 9 t j
p − 3 = − 12 , q − 12 = − 9. p-3=-12,\qquad q-12=-9. p − 3 = − 12 , q − 12 = − 9. Thus p = − 9 , q = 3 \boxed{p=-9,\quad q=3} p = − 9 , q = 3
(c)
解法一
思路
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先用距离等于 15 15 15 B B B A A A
答题过程
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When the distance is 15 km,
( 25 − 12 t ) 2 + ( − 9 t ) 2 = 15 2 , 225 t 2 − 600 t + 400 = 0 , ( 15 t − 20 ) 2 = 0 , t = 4 3 h . \begin{align*}
(25-12t)^2+(-9t)^2=&\,15^2,\\
225t^2-600t+400=&\,0,\\
(15t-20)^2=&\,0,\\
t=&\,\frac43\ \mathrm h.
\end{align*} ( 25 − 12 t ) 2 + ( − 9 t ) 2 = 225 t 2 − 600 t + 400 = ( 15 t − 20 ) 2 = t = 1 5 2 , 0 , 0 , 3 4 h . At this time,
A B → = 9 i − 12 j k m , B A → = − 9 i + 12 j k m . \overrightarrow{AB}=9\mathbf i-12\mathbf j\ \mathrm{km},\qquad
\overrightarrow{BA}=-9\mathbf i+12\mathbf j\ \mathrm{km}. A B = 9 i − 12 j km , B A = − 9 i + 12 j km . Thus A A A B B B θ \theta θ
tan θ = 9 12 , θ = 36.8699 … ∘ . \tan\theta=\frac9{12},\qquad \theta=36.8699\ldots^\circ. tan θ = 12 9 , θ = 36.8699 … ∘ . The bearing of A A A B B B
360 ∘ − θ = 323.130 … ∘ = 323 ∘ 360^\circ-\theta=323.130\ldots^\circ=\boxed{323^\circ} 36 0 ∘ − θ = 323.130 … ∘ = 32 3 ∘ to the nearest degree.
