Question

A boat of mass m = 100 kg is being tugged at a constant velocity $v_0$ = 3 m/s on a lake by a tugboat with the help of a rope. How far the boat will move after the rope is cut. Force $\bar{f}$ of water drag depends on velocity $\bar{v}$ and acceleration $\bar{a}$ of the boat according to the law $\bar{f} = -\alpha\bar{v} + \beta\bar{a}$, where $\alpha = 10$ N⋅s/m and $\beta$ = 50 N⋅s²/m.

Answer 1

15 m

Answer 2

The problem involves a boat decelerating due to a velocity-dependent and acceleration-dependent drag force after its rope is cut. We need to find the total distance it travels until it stops.

1. Set up the Equation of Motion: After the rope is cut, the only force acting on the boat is the water drag force $\bar{f}$. According to Newton's second law, the net force equals mass times acceleration: $m\bar{a} = \bar{f}$

Given the drag force law $\bar{f} = -\alpha\bar{v} + \beta\bar{a}$, we substitute it into the equation of motion: $m\bar{a} = -\alpha\bar{v} + \beta\bar{a}$

Rearrange the terms to group acceleration: $m\bar{a} - \beta\bar{a} = -\alpha\bar{v}$ $(m - \beta)\bar{a} = -\alpha\bar{v}$

Let's consider the magnitudes for one-dimensional motion, where $\bar{v}$ and $\bar{a}$ are in the same direction (or opposite for deceleration). $(m - \beta)a = -\alpha v$

2. Relate Acceleration, Velocity, and Position: To find the distance, it's convenient to express acceleration $a$ as $v \frac{dv}{dx}$, where $v$ is velocity and $x$ is position. $(m - \beta)v \frac{dv}{dx} = -\alpha v$

3. Solve the Differential Equation: Since the boat is moving (i.e., $v \neq 0$) for the duration of its travel until it stops, we can divide both sides by $v$: $(m - \beta)\frac{dv}{dx} = -\alpha$

Now, separate variables and integrate. The velocity changes from the initial velocity $v_0$ to $0$ (when the boat stops), and the position changes from $0$ to the final distance $X$: $\int_{v_0}^{0} (m - \beta)dv = \int_{0}^{X} -\alpha dx$

Integrate both sides: $(m - \beta)[v]_{v_0}^{0} = -\alpha[x]_{0}^{X}$ $(m - \beta)(0 - v_0) = -\alpha(X - 0)$ $-(m - \beta)v_0 = -\alpha X$

4. Calculate the Distance X: Solve for $X$: $X = \frac{(m - \beta)v_0}{\alpha}$

Substitute the given values: $m = 100 \text{ kg}$ $\beta = 50 \text{ N}\cdot\text{s}^2/\text{m}$ (Note: $1 \text{ N}\cdot\text{s}^2/\text{m} = (1 \text{ kg}\cdot\text{m/s}^2)\cdot\text{s}^2/\text{m} = 1 \text{ kg}$, so $\beta$ effectively acts as a mass term) $v_0 = 3 \text{ m/s}$ $\alpha = 10 \text{ N}\cdot\text{s/m}$

$X = \frac{(100 \text{ kg} - 50 \text{ kg}) \times 3 \text{ m/s}}{10 \text{ N}\cdot\text{s/m}}$ $X = \frac{(50 \text{ kg}) \times 3 \text{ m/s}}{10 \text{ N}\cdot\text{s/m}}$ $X = \frac{150 \text{ kg}\cdot\text{m/s}}{10 \text{ kg/s}}$ $X = 15 \text{ m}$

The boat will move 15 meters after the rope is cut.