一个在某种介质中运动的物体，速度为 $v$ m/s，受到大小为 $R$ 牛顿的运动阻力。$R$ 与 $v$ 满足方程 $R = av^2 + bv$，其中 $a$ 和 $b$ 为常数。 下表显示了变量 $v$ 和 $R$ 的实验测量值。

$v$	5	10	15	20	25
$R$	17	44	81	128	185

1. Calculate the values of $\frac{R}{v}$ for each corresponding $v$.

For $v = 5$, $\frac{R}{v} = \frac{17}{5} = 3.4$
For $v = 10$, $\frac{R}{v} = \frac{44}{10} = 4.4$
For $v = 15$, $\frac{R}{v} = \frac{81}{15} = 5.4$
For $v = 20$, $\frac{R}{v} = \frac{128}{20} = 6.4$
For $v = 25$, $\frac{R}{v} = \frac{185}{25} = 7.4$

2. Plot the points on the grid and draw the straight line.

Choose suitable scales for the $v$-axis (horizontal) and $\frac{R}{v}$-axis (vertical).
Plot the points $(5, 3.4)$, $(10, 4.4)$, $(15, 5.4)$, $(20, 6.4)$, and $(25, 7.4)$.
Draw a straight line passing through all these points.

1. Express the given equation in linear form

Given $R = av^2 + bv$, dividing by $v$ gives:
$\frac{R}{v} = av + b$
This is in the form $Y = mX + c$, where $Y = \frac{R}{v}$, $X = v$, gradient $m = a$, and vertical intercept $c = b$.

2. Calculate the gradient to find $a$

Using two points from the line, e.g., $(5, 3.4)$ and $(25, 7.4)$:
$a = \text{gradient} = \frac{7.4 - 3.4}{25 - 5} = \frac{4}{20} = 0.2$

3. Find the vertical intercept to determine $b$

From the graph, the vertical intercept (where $v = 0$) is $2.4$.
Alternatively, using a point $(5, 3.4)$:
$3.4 = 0.2(5) + b \implies b = 3.4 - 1 = 2.4$

1. Find the equation relating $R$ and $v$ for the second medium.

Since $R$ is directly proportional to $v$, $R = kv$.
Given $R = 30$ when $v = 5$:
$30 = k(5) \implies k = 6$
So, $R = 6v$.

2. Determine the equation of the line to be drawn on the $\frac{R}{v}$ against $v$ graph.

Divide the equation by $v$ to match the $y$-axis of the graph:
$\frac{R}{v} = 6$
This is a horizontal line passing through $6$ on the vertical axis.

3. Find the value of $v$ where the resistance is the same.

The resistance is the same when the two graphs intersect.
From the graph of $\frac{R}{v}$ against $v$, the line $\frac{R}{v} = 6$ intersects the original line at $v = 18$.