题目
Problem
7.
[In this question i \mathbf{i} i j \mathbf{j} j
Two boats, A A A B B B
At time t t t A A A v A k m h − 1 \mathbf{v}_A\,\mathrm{km\,h^{-1}} v A km h − 1
v A = 2 i + 3 j + ( i − 4 j ) t \mathbf{v}_A=2\mathbf{i}+3\mathbf{j}+(\mathbf{i}-4\mathbf{j})t v A = 2 i + 3 j + ( i − 4 j ) t (a) Find the magnitude of the acceleration of A A A
(2)
When t = 2 t=2 t = 2 B B B ( 4 i + j ) k m h − 1 (4\mathbf{i}+\mathbf{j})\,\mathrm{km\,h^{-1}} ( 4 i + j ) km h − 1
When t = 5 t=5 t = 5 B B B ( i − 5 j ) k m h − 1 (\mathbf{i}-5\mathbf{j})\,\mathrm{km\,h^{-1}} ( i − 5 j ) km h − 1
(b) Find the acceleration of B B B i \mathbf{i} i j \mathbf{j} j
(2)
(c) Find the velocity of B B B t = 0 t=0 t = 0 i \mathbf{i} i j \mathbf{j} j
(2)
At time T 1 T_1 T 1
(d) Find the exact value of T 1 T_1 T 1
(4)
At time T 2 T_2 T 2
(e) Show that 3 T 2 2 + p T 2 + q = 0 3T_2^2+pT_2+q=0 3 T 2 2 + p T 2 + q = 0 p p p q q q
(3)
题目中文翻译
[本题中，i \mathbf{i} i j \mathbf{j} j
两艘船 A A A B B B
正午之后 t t t A A A v A k m h − 1 \mathbf{v}_A\,\mathrm{km\,h^{-1}} v A km h − 1
v A = 2 i + 3 j + ( i − 4 j ) t \mathbf{v}_A=2\mathbf{i}+3\mathbf{j}+(\mathbf{i}-4\mathbf{j})t v A = 2 i + 3 j + ( i − 4 j ) t (a) 求 A A A
(2)
当 t = 2 t=2 t = 2 B B B ( 4 i + j ) k m h − 1 (4\mathbf{i}+\mathbf{j})\,\mathrm{km\,h^{-1}} ( 4 i + j ) km h − 1
当 t = 5 t=5 t = 5 B B B ( i − 5 j ) k m h − 1 (\mathbf{i}-5\mathbf{j})\,\mathrm{km\,h^{-1}} ( i − 5 j ) km h − 1
(b) 求 B B B i \mathbf{i} i j \mathbf{j} j
(2)
(c) 求 t = 0 t=0 t = 0 B B B i \mathbf{i} i j \mathbf{j} j
(2)
正午之后 T 1 T_1 T 1
(d) 求 T 1 T_1 T 1
(4)
正午之后 T 2 T_2 T 2
(e) 证明 3 T 2 2 + p T 2 + q = 0 3T_2^2+pT_2+q=0 3 T 2 2 + p T 2 + q = 0 p p p q q q
(3)
解答
(a)
解法一
思路
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恒加速度就是速度表达式中 t t t
答题过程
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The acceleration of A A A
a A = i − 4 j k m h − 2 . \mathbf a_A=\mathbf i-4\mathbf j\ \mathrm{km\,h^{-2}}. a A = i − 4 j km h − 2 . Its magnitude is
∣ a A ∣ = 1 2 + ( − 4 ) 2 = 17 k m h − 2 . |\mathbf a_A|=\sqrt{1^2+(-4)^2}=\boxed{\sqrt{17}\ \mathrm{km\,h^{-2}}}. ∣ a A ∣ = 1 2 + ( − 4 ) 2 = 17 km h − 2 .
(b)
解法一
思路
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用两时刻速度之差除以时间差 3 小时，最终保留向量形式。
答题过程
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The constant acceleration of B B B
a B = ( i − 5 j ) − ( 4 i + j ) 5 − 2 , = ( − i − 2 j ) k m h − 2 . \begin{align*}
\mathbf a_B=&\,\frac{(\mathbf i-5\mathbf j)-(4\mathbf i+\mathbf j)}{5-2},\\
=&\,\boxed{(-\mathbf i-2\mathbf j)\ \mathrm{km\,h^{-2}}}.
\end{align*} a B = = 5 − 2 ( i − 5 j ) − ( 4 i + j ) , ( − i − 2 j ) km h − 2 .
(c)
解法一
思路
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从第 2 小时的速度倒推两小时，初速度等于该速度减去 2 a B 2\mathbf a_B 2 a B
答题过程
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Using v B = u B + t a B \mathbf v_B=\mathbf u_B+t\mathbf a_B v B = u B + t a B t = 2 t=2 t = 2
u B = ( 4 i + j ) − 2 ( − i − 2 j ) , = ( 6 i + 5 j ) k m h − 1 . \begin{align*}
\mathbf u_B=&\,(4\mathbf i+\mathbf j)-2(-\mathbf i-2\mathbf j),\\
=&\,\boxed{(6\mathbf i+5\mathbf j)\ \mathrm{km\,h^{-1}}}.
\end{align*} u B = = ( 4 i + j ) − 2 ( − i − 2 j ) , ( 6 i + 5 j ) km h − 1 .
(d)
解法一
思路
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速率相等只要求速度向量的模相等，不要求两分量相同。把模平方相等展开，解二次式后用正午之后的时间条件选正根。
答题过程
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At time t t t
v A = ( 2 + t ) i + ( 3 − 4 t ) j , v B = ( 6 − t ) i + ( 5 − 2 t ) j . \mathbf v_A=(2+t)\mathbf i+(3-4t)\mathbf j,\qquad
\mathbf v_B=(6-t)\mathbf i+(5-2t)\mathbf j. v A = ( 2 + t ) i + ( 3 − 4 t ) j , v B = ( 6 − t ) i + ( 5 − 2 t ) j . For equal speeds at t = T 1 t=T_1 t = T 1
( 6 − T 1 ) 2 + ( 5 − 2 T 1 ) 2 = ( 2 + T 1 ) 2 + ( 3 − 4 T 1 ) 2 . (6-T_1)^2+(5-2T_1)^2
=(2+T_1)^2+(3-4T_1)^2. ( 6 − T 1 ) 2 + ( 5 − 2 T 1 ) 2 = ( 2 + T 1 ) 2 + ( 3 − 4 T 1 ) 2 . Expanding both sides,
5 T 1 2 − 32 T 1 + 61 = 17 T 1 2 − 20 T 1 + 13 , 12 T 1 2 + 12 T 1 − 48 = 0 , T 1 2 + T 1 − 4 = 0. \begin{align*}
5T_1^2-32T_1+61=&\,17T_1^2-20T_1+13,\\
12T_1^2+12T_1-48=&\,0,\\
T_1^2+T_1-4=&\,0.
\end{align*} 5 T 1 2 − 32 T 1 + 61 = 12 T 1 2 + 12 T 1 − 48 = T 1 2 + T 1 − 4 = 17 T 1 2 − 20 T 1 + 13 , 0 , 0. Thus T 1 = ( − 1 ± 17 ) / 2 T_1=(-1\pm\sqrt{17})/2 T 1 = ( − 1 ± 17 ) /2 T 1 > 0 T_1>0 T 1 > 0
T 1 = 17 − 1 2 . \boxed{T_1=\frac{\sqrt{17}-1}{2}}. T 1 = 2 17 − 1 .
(e)
解法一
思路
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同方向要求速度向量为正倍数，分量比必相等。先交叉相乘并完整展开，再与题目指定的形式比较系数。
答题过程
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For the velocities to have the same direction at t = T 2 t=T_2 t = T 2
5 − 2 T 2 6 − T 2 = 3 − 4 T 2 2 + T 2 . \frac{5-2T_2}{6-T_2}=\frac{3-4T_2}{2+T_2}. 6 − T 2 5 − 2 T 2 = 2 + T 2 3 − 4 T 2 . Cross-multiplying and expanding,
( 5 − 2 T 2 ) ( 2 + T 2 ) = ( 3 − 4 T 2 ) ( 6 − T 2 ) , 10 + T 2 − 2 T 2 2 = 18 − 27 T 2 + 4 T 2 2 , 6 T 2 2 − 28 T 2 + 8 = 0. \begin{align*}
(5-2T_2)(2+T_2)=&\,(3-4T_2)(6-T_2),\\
10+T_2-2T_2^2=&\,18-27T_2+4T_2^2,\\
6T_2^2-28T_2+8=&\,0.
\end{align*} ( 5 − 2 T 2 ) ( 2 + T 2 ) = 10 + T 2 − 2 T 2 2 = 6 T 2 2 − 28 T 2 + 8 = ( 3 − 4 T 2 ) ( 6 − T 2 ) , 18 − 27 T 2 + 4 T 2 2 , 0. Dividing by 2 gives
3 T 2 2 − 14 T 2 + 4 = 0 , \boxed{3T_2^2-14T_2+4=0}, 3 T 2 2 − 14 T 2 + 4 = 0 , so p = − 14 , q = 4 \boxed{p=-14,\ q=4} p = − 14 , q = 4
The two roots are ( 7 ± 37 ) / 3 (7\pm\sqrt{37})/3 ( 7 ± 37 ) /3 i \mathbf i i 2 + T 2 2+T_2 2 + T 2 6 − T 2 6-T_2 6 − T 2
