Problem 75 - Physics for Scientists and Engineers with Modern Physics

Question

(III) A motorboat traveling at a speed of \(2.4 \mathrm{~m} / \mathrm{s}\) shuts off its engines at \(t=0 .\) How far does it travel before coming to rest if it is noted that after \(3.0 \mathrm{~s}\) its speed has dropped to half its original value? Assume that the drag force of the water is proportional to \(v\)

Step-by-Step Solution

Answer

The motorboat travels approximately 10.39 meters before coming to rest.

Step 1: Define Initial Conditions and Given Values

We are given that the initial speed of the motorboat is \( v_0 = 2.4 \, \text{m/s} \). At \( t = 3.0 \, \text{s} \), the speed is half the initial value, which is \( v = 1.2 \, \text{m/s} \). The drag force is proportional to speed, indicating a linear deceleration model.

Step 2: Determine Acceleration Function

Given the drag force is proportional to speed, use the relationship \( F_d = -kv \) where \( k \) is a constant. Applying Newton's second law \( F = ma \), we have \( ma = -kv \). Therefore, \( a = \frac{-k}{m} v \), leading to exponential decay in speed described by \( v(t) = v_0 e^{-\frac{k}{m}t} \).

Step 3: Solve for the Decay Constant \( k/m \)

Using the information that \( v(3) = \frac{v_0}{2} \), substitute into the velocity equation: \[ \frac{v_0}{2} = v_0 e^{-\frac{k}{m} \cdot 3} \]. Simplifying gives \[ \frac{1}{2} = e^{-\frac{k}{m} \cdot 3} \]. Taking the natural logarithm of both sides, we find \[ -\frac{k}{m} \cdot 3 = \ln\left(\frac{1}{2}\right) \] so \[ \frac{k}{m} = \frac{\ln\left(\frac{1}{2}\right)}{-3} \].

Step 4: Integrate the Velocity Function to Find Distance

The displacement \( s(t) \) is given by the integration of velocity over time: \[ s(t) = \int_0^t v_0 e^{-\frac{k}{m} \tau} d\tau \]. Substitute \( v_0 = 2.4 \) m/s and \( \frac{k}{m} = \frac{-\ln\left(\frac{1}{2}\right)}{3} \) to find the integral \[ s(t) = v_0 \left[-\frac{m}{k} e^{-\frac{k}{m}t} \right]_0^t = \frac{2.4m}{k} \left(1 - e^{-\frac{k}{m}t}\right) \].

Step 5: Evaluate the Total Distance Until Rest

The boat comes to rest when its velocity is zero, which occurs as \( t \to \infty \). Therefore, the total distance \( s_{\infty} \) when it's at rest is: \[ s_{\infty} = \frac{2.4m}{k} \]. Using \( \frac{k}{m} = \frac{-\ln\left(\frac{1}{2}\right)}{3} \), \( \frac{m}{k} = -\frac{3}{\ln\left(\frac{1}{2}\right)} \). Substitute into the distance equation: \[ s_{\infty} = 2.4 \times -\frac{3}{\ln\left(\frac{1}{2}\right)} \approx 10.39 \, \text{meters}. \]

Key Concepts

Newton's Second Law

Newton's Second Law forms the foundation for understanding dynamics and motion. It states that the force acting on an object is equal to the mass of the object multiplied by the acceleration it experiences, expressed as \( F = ma \). In the context of the motorboat problem, the drag force imposed by the water, which slows the boat down, can be described using this law.

When the engines of the motorboat are off, the only significant horizontal force is the drag force. As specified, this drag force is proportional to the speed of the motorboat, and can be expressed as \( F_d = -kv \), where \( k \) is a proportionality constant and \( v \) is the velocity. The negative sign indicates that the force opposes the motion.

• The mass \( m \) of the boat is a crucial factor—it determines how much the velocity changes over time due to the drag force.
• Newton's second law allows us to equate the drag force to the product of mass and acceleration, giving us \( ma = -kv \).

This relationship helps us determine how the velocity of the boat changes with time, forming the basis for calculating the total distance traveled until the boat comes to rest.

Exponential Decay

Exponential decay is a process where a quantity decreases at a rate proportional to its current value. In our motorboat problem, the speed decreases over time following an exponential decay model. This is because the acceleration, derived from Newton's second law, is also proportional to the velocity with a negative sign, as seen in \( a = \frac{-k}{m} v \).

Given the initial velocity \( v_0 \) of the motorboat at time \( t = 0 \), the velocity \( v(t) \) at any time \( t \) is described by the equation:
\[ v(t) = v_0 e^{-\frac{k}{m}t} \]

• This equation shows that the speed decreases exponentially as time passes.
• The negative exponent indicates that the velocity diminishes over time, with the rate of change depending on the constant \( \frac{k}{m} \).

Given a specific time, such as when the speed is half the initial value, we can solve for \( \frac{k}{m} \) using the velocity equation to quantify how quickly the speed decays. This factor is crucial for understanding how long and far the boat will travel before stopping.

Linear Deceleration

Linear deceleration may sound contradictory in a context describing exponential decay, but it points to the manner in which the deceleration depends directly on the velocity, making it a linear function of \( v \).

In the motorboat scenario, we understand this as the boat slowing down at a rate that's directly tied to how fast it's moving at any point in time. If the speed is high, the drag force (and thus deceleration) is also high, represented by \( F_d = -kv \).

• The concept of linear deceleration helps us focus on the direct proportionality between the speed and the drag force.
• Deceleration, or the rate of slowing down, is thus constantly decreasing as the boat slows.

The formulation of \( a = \frac{-k}{m} v \) highlights that although the speed undergoes exponential decay, the model used for the deceleration can be seen as "linear" with respect to the instantaneous velocity. The term serves to simplify the understanding of how the principal force involved, drag, scales linearly with speed, ensuring a seamless understanding of the motion until the motorboat stops.