Solution 6
We need to determine the maximum height $h$ the robot can climb. The height climbed depends on the speed $v$ and the battery life $\tau$.
1. Formulation
The robot climbs at a constant speed $v$ for a duration $\tau$. The total height is:
$$ h = v \times \tau $$The graph provides the relationship between the reciprocal of time $(1/\tau)$ and speed $v$. Let $y = 1/\tau$. Substituting $\tau = 1/y$ into the height equation:
$$ h = \frac{v}{y} $$2. Graphical Interpretation
To maximize $h = v/y$, we need to find the point on the curve where the ratio of the x-coordinate ($v$) to the y-coordinate ($y$) is maximum.
Geometrically, $y/v$ represents the slope of a line connecting the origin $(0,0)$ to a point $(v, y)$ on the curve. Therefore, $v/y$ is the inverse of this slope. To maximize the inverse slope, we must minimize the slope of the line drawn from the origin to the curve.
This minimum slope corresponds to the tangent drawn from the origin to the curve.
Figure 2: Plot of reciprocal discharge time vs. speed. The red dashed line indicates the tangent from the origin, representing the condition for maximum range.
3. Calculation
By inspecting the graph and identifying the point of tangency, we find the operating point coordinates:
- Velocity $v = 2.0 \, \text{m/s}$
- Reciprocal time $1/\tau = 0.04 \, \text{s}^{-1}$
Now we calculate the maximum height:
$$ h_{max} = \frac{v}{(1/\tau)} = \frac{2.0}{0.04} $$ $$ h_{max} = \frac{200}{4} = 50 \, \text{m} $$Answer: The maximum length of the staircase the robot can climb is 50 m.