《2025年全程检测单元测试卷九年级数学全一册湘教版》

17. (12分)某汽车的功率P(W)为定值,它的速度v(m/s)与它所受的牵引力F(N)之间关系的函数表达式为$ v = \frac{P}{F} $,且当$ F = 4000N $时,$ v = 20m/s $.
(1)(6分)这辆汽车的功率是多少?请写出这个函数的表达式.
(2)(6分)当它所受的牵引力为3200N时,汽车的速度为多少?

答案： 17.
(1) 由题意得$v = \frac{P}{F}$，这辆汽车的功率$P = F \cdot v = 4000 × 20 = 80000(W)$，所求的函数表达式为$v = \frac{80000}{F}$.
(2) 当$F = 3200 N$时，$v = \frac{80000}{3200} = 25(m/s)$，$\therefore$汽车的速度为$25 m/s$.

18. (12分)如图7,A,C分别是x轴、y轴上的点,且$ OC = 2 $,$ \angle CAO = 30° $,延长AC交反比例函数$ y = \frac{k}{x}(x ＞ 0) $的图象于点B,过点B作$ BD \perp y $轴于点D,已知$ OC:CD = 2:1 $.
(1)(6分)求这个反比例函数的表达式.
(2)(6分)求$ \triangle AOB $的面积.

答案： 18.
(1) $\because OC = 2,OC:CD = 2:1,\therefore CD = 1,OD = 3.\because BD \perp y$轴，$\therefore BD // x$轴，$\angle BDC = 90^{\circ}.\therefore \angle DBC = \angle CAO = 30^{\circ}.\therefore BC = 2CD = 2$. 由勾股定理得$DB = \sqrt{BC^2 - CD^2} = \sqrt{2^2 - 1^2} = \sqrt{3}$，$\therefore$点$B$的坐标为$(\sqrt{3},3)$. 把点$B(\sqrt{3},3)$的坐标代入$y = \frac{k}{x}$，得$k = 3\sqrt{3}.\therefore$所求的反比例函数表达式为$y = \frac{3\sqrt{3}}{x}$.
(2) $\because$在$Rt \triangle ACO$中，$OC = 2,\angle CAO = 30^{\circ},\therefore AC = 2OC = 4$. 由勾股定理，得$OA = \sqrt{AC^2 - OC^2} = \sqrt{4^2 - 2^2} = 2\sqrt{3}$，又$\because$点$B$的坐标为$(\sqrt{3},3)$，$\therefore \triangle AOB$的面积为$\frac{1}{2} × 2\sqrt{3} × 3 = 3\sqrt{3}$.