Question

A motorboat moves in still water with a speed \(\mathrm{v}=10 \mathrm{~km} / \mathrm{h}\). At full speed its engine was cut off and in 20 seconds the speed was reduced to \(\mathrm{v}_{1}=6 \mathrm{~km} / \mathrm{h}\). Assuming that the force of water resistance to the moving boat is proportional to its speed, find the speed of the boat in two minutes after the engine was shut off; find also the distance travelled by the boat during one minute with the engine dead.

Short answer

Answer: The boat's approximate speed after two minutes with the engine off is 6.65 km/h, and it travels approximately 93 meters during the first minute with the engine off.

Step-by-step solution

1. Determine the exponential decay formula:

We will use the exponential decay formula: \(v(t) = v_0 \times (e^{-kt})\), where \(v_0\) is the initial speed (in km/h), \(v(t)\) is the speed at time \(t\) (in hours), and \(k\) is the decay constant.

2. Convert the given speed and time data to base units:

First, convert 20 seconds to hours: \(\frac{20}{3600} = \frac{1}{180}\) hours. Then, convert the speed to base units: \(v_0 = 10\) km/h, \(v_1 = 6\) km/h.

3. Calculate the decay constant \(k\):

We will use the exponential decay formula with the given values at \(t = \frac{1}{180}\) h: \(6 = 10(e^{-k(\frac{1}{180})})\). Solve for \(k\): \(\frac{6}{10} = e^{-\frac{k}{180}} \Rightarrow k = -180 \ln(\frac{3}{5})\).

4. Find the speed after two minutes:

Next, we will find the speed after two minutes (t = 2/60 hours = 1/30 hours): \(v(\frac{1}{30}) = 10 \times e^{-\frac{1}{30} \times (-180 \ln(\frac{3}{5}))}\).

5. Calculate the speed of the boat:

Use a calculator to find the speed after two minutes: \(v(\frac{1}{30}) = 10 \times e^{\frac{1}{30} \times (180 \ln(\frac{3}{5}))} \approx 6.65\) km/h.

6. Find the distance traveled during one minute:

We will integrate the exponential decay formula to find the distance traveled in the first minute (from \(t = 0\) to \(t = \frac{1}{60}\) hour): \(s(\frac{1}{60}) = \int_{0}^{\frac{1}{60}}10e^{-t(-180 \ln(\frac{3}{5}))}dt\).

7. Calculate the distance traveled in one minute:

Using a calculator or software to integrate and evaluate the integral: \(s(\frac{1}{60}) = \left[-\frac{50}{3\ln{(\frac{5}{3})}}e^{-t(180 \ln(\frac{3}{5}))}\right ]_{0}^{\frac{1}{60}} \approx 0.093\) km (93 meters). The speed of the boat after two minutes with the engine off is approximately 6.65 km/h and the distance traveled by the boat during the first minute with the engine off is approximately 93 meters.

Key concepts

Differential Equations

Differential equations are mathematical tools that describe the relationship between a function and its derivatives. They are essential for modeling the change of various quantities over time, particularly in the natural sciences and engineering. When a differential equation involves a single independent variable, we often seek a function which, after being substituted into the equation, will satisfy it for all values of the variable.

In the context of the motorboat problem provided, a differential equation is used to model the decay of speed over time. The force of water resistance, which is proportional to the boat's speed, influences this decay, leading us to an exponential decrease in speed. Let's consider the function v(t) representing the speed of the motorboat at time t. A differential equation governing the boat's speed might look something like dv/dt = -k * v, where k is the positive decay constant. The negative sign indicates a decrease in speed.

When solving such an exponential decay problem, we start with the general form of the differential equation and apply initial conditions to solve for any constants. The solution process typically involves separating variables and integrating both sides of the equation—a perfect segue to our next concept, integration calculus.

Natural Logarithm

Understanding the Natural Logarithm

The natural logarithm, denoted as ln, is the logarithm to the base e, where e is an irrational and transcendental number approximately equal to 2.71828. This special logarithm is intrinsic to exponential growth and decay processes, appearing frequently in problems involving continuous growth rates, such as population growth, radioactive decay, and yes, the slowing down of a motorboat in water.

In our exercise, we solve for the decay constant k using the natural logarithm. It helps us to unpack the exponent in the equation v(t) = v_0 * (e^{-kt}). When we have 6/10 = e^{-k/180} and wish to isolate k, we apply the natural logarithm to both sides, obtaining k = -180 * ln(3/5). The natural logarithm is thus crucial for solving exponential equations, allowing us to manipulate and solve for variables that are located in the exponent.

Integration Calculus

Applying Integration to Calculate Distance

Integration calculus is the reverse process of differentiation and is used for finding areas, volumes, central points and many useful things. Essentially, integration can be used to add up slices to find the whole. Integration is a core component of calculus, commonly symbolized by the integral sign ∫ followed by the function to be integrated and the differential.

In the given problem, we use integration to find the distance traveled by the boat over a specific time interval. The distance s(t) traveled from time t = 0 to time t is the integral of the speed function with respect to time. For the exponential speed function v(t) = 10e^{-k*t}, integrating with respect to time from 0 to 1/60 hours yields the total distance traveled in the first minute after the engine cuts off. The actual integration, which may involve techniques like u-substitution or integration by parts, gives us the precise value for this distance.

• Integration transforms a speed-time graph into a distance.
• It involves finding the area under the speed-time curve over the given interval.
• The result of this particular integration gives the distance the boat has traveled in one minute.