NLM O29

1. Analyzing the Force Law

From the provided graph, we observe a linear relationship between the resistive force $F$ and velocity $v$. The line passes through $v=0$ where the force becomes zero. This implies a direct proportionality:

$$ F = -kv $$

where $k$ is a constant coefficient of viscous drag. The negative sign indicates the force opposes motion.

2. Setting up the Equation of Motion

Using Newton’s Second Law ($F = ma$):

$$ m a = -kv $$

We can rewrite acceleration $a$ as $v \frac{dv}{dx}$ to relate velocity to distance:

$$ m \left( v \frac{dv}{dx} \right) = -kv $$

Canceling $v$ from both sides (for $v \neq 0$):

$$ m \frac{dv}{dx} = -k $$ $$ dv = -\frac{k}{m} dx $$

3. Integrating to find Stopping Distance

Let the particle travel from an initial velocity $v_0$ to a final velocity $0$ over a distance $S$. Integrating both sides:

$$ \int_{v_0}^{0} dv = -\frac{k}{m} \int_{0}^{S} dx $$ $$ [v]_{v_0}^{0} = -\frac{k}{m} [x]_{0}^{S} $$ $$ 0 – v_0 = -\frac{k}{m} S $$ $$ v_0 = \frac{k}{m} S \implies S = \left( \frac{m}{k} \right) v_0 $$

This shows that the stopping distance $S$ is directly proportional to the initial velocity $v_0$.

$$ S \propto v_0 $$

4. Calculation

We are given a reference case and asked to find the distance for a new case.

• Case 1: Initial velocity $v_1 = 8 \text{ m/s}$, Stopping distance $S_1 = 20 \text{ m}$.
• Case 2: Initial velocity $v_2 = 20 \text{ m/s}$ (from the very beginning), Stopping distance $S_2 = ?$.

Using the proportionality $S \propto v_0$:

$$ \frac{S_2}{S_1} = \frac{v_2}{v_1} $$ $$ S_2 = S_1 \times \frac{v_2}{v_1} $$ $$ S_2 = 20 \text{ m} \times \frac{20 \text{ m/s}}{8 \text{ m/s}} $$ $$ S_2 = 20 \times 2.5 = 50 \text{ m} $$