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Chapter 6: Problem 41

A \(1000 \mathrm{~kg}\) boat is traveling at \(90 \mathrm{~km} / \mathrm{h}\) when its engine is shut off. The magnitude of the frictional force \(\vec{f}^{\text {kin }}\) between boat and water is proportional to the speed \(v\) of the boat; \(\vec{f}^{\mathrm{kin}}=(70 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}) \vec{v}\). Find the time required for the boat to slow to \(45 \mathrm{~km} / \mathrm{h}\).

Short Answer

Expert verified
The time required for the boat to slow to 45 km/h is approximately 9.9 seconds.

Step by step solution

01

Convert speeds to meters per second

First, convert the initial speed and final speed from km/h to m/s using the conversion factor: 1 km/h = \frac{1}{3.6}\ m/s. Therefore, Initial speed: \(v_0 = 90 \frac{km}{h} \times \frac{1}{3.6}\ = 25 \frac{m}{s} \)Final speed: \( v_f = 45 \frac{km}{h} \times \frac{1}{3.6}\ = 12.5 \frac{m}{s} \)
02

Write the acceleration using Newton's second law

Given the frictional force as \( \vec{f}^{\mathrm{kin}} = (70 \frac{N \cdot s}{m}) \vec{v} \), and using Newton's second law (\( \vec{F} = m \vec{a} \): \The friction force is given by: \( f = -70v \)According to the second law: \( f = m \cdot a \)thus \( -70v = 1000 \cdot a \)and therefore \( a = -0.07v \)
03

Set up the differential equation

The acceleration \(\ a \) is the derivative of velocity, i.e., \( a = \frac{dv}{dt} \). Substitute \( a = -0.07v \) into the equation: \( \frac{dv}{dt} = -0.07v \)
04

Solve the differential equation

To solve \( \frac{dv}{dt} = -0.07v \), separate the variables: \( \frac{dv}{v} = -0.07 dt \)Integrate both sides: \( \int \frac{1}{v} dv = -0.07 \int dt \)This gives: \( \ln|v| = -0.07t + C \)where \(C\) is the integration constant. Solve for \(v\): \( v = e^{-0.07t + C} \)Or: \( v = e^C \cdot e^{-0.07t} \)Let \( A = e^C \), so \( v = A e^{-0.07 t} \)
05

Determine the integration constant

To find the constant \( A \), use the initial condition \( v_0 = 25 \frac{m}{s} \): \( 25 = A e^{-0.07(0)} \)Therefore: \( A = 25 \)
06

Solve for time

Now that the velocity function is \( v = 25 e^{-0.07t} \), find the time \( t \) when the velocity \( v_f = 12.5 \frac{m}{s} \): \( 12.5 = 25 e^{-0.07t} \)Divide both sides by 25: \( 0.5 = e^{-0.07 t} \)Take the natural logarithm of both sides: \( \ln(0.5) = -0.07 t \)So: \( t = \frac{\ln(0.5)}{-0.07} \)Calculate the final result: \( t \approx \frac{-0.693}{-0.07} \approx 9.9 \text{s} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conversion of Units
To solve problems involving physics, correct unit conversion is crucial. In the problem, speeds are given in kilometers per hour (km/h). But in physics equations, we often need meters per second (m/s). To convert from km/h to m/s, we use the conversion factor: 1 km/h = \( \frac{1}{3.6} \ m/s \).
For example:
  • Initial speed: \(90 \ km/h \times \frac{1}{3.6} = 25 \ m/s \)
Ensure you check your conversions to avoid mistakes. This will help in accurately solving the next steps.
Frictional Force
Frictional force plays a major role in slowing down objects moving through a medium like water. The problem states that the frictional force \( \vec{f}^{\text{kin}} \) is proportional to the speed \( v \) of the boat, represented by:
\( \vec{f}^{\text{kin}}=(70 \ N \cdot \ s/m) \cdot \vec{v} \)
This means the faster the boat goes, the larger the frictional force. This relationship causes the boat to decelerate over time as it slows down, ultimately leading it to stop, given a long enough timeframe.
Differential Equations
Differential equations are equations that involve derivatives. In this problem, Newton's second law and the given frictional force enable us to set up a differential equation. Using Newton's Second Law \( \vec{F} = m \cdot \vec{a} \):
We combine this with \( \vec{f}^{\text{kin}} = -70 v \) and acceleration \( \vec{a} = \frac{d \vec{v}}{dt} \) gives:
\[ \frac{d \vec{v}}{dt} = -0.07 \vec{v} \]
This is a first-order linear differential equation and solving it involves integrating both sides to find the velocity function over time.
Exponential Decay
Exponential decay refers to the decrease in a quantity at a rate proportional to its current value. Here, the boat's velocity decreases exponentially over time. By solving the differential equation \( \frac{d \vec{v}}{dt} = -0.07 \vec{v} \), we get:
\[ \vec{v} = A e^{-0.07t} \]
  • A is the initial speed; in this case, it is 25 m/s.
This decay function tells us how the boat鈥檚 speed decreases exponentially as time progresses, becoming slower and slower.
Velocity and Acceleration
Understanding velocity and acceleration is fundamental in this problem. Velocity (\( v \)) is the speed of the boat in a defined direction. Acceleration (\( a \)) is the rate of change of velocity. In the problem:
\[ a = \frac{dv}{dt} = -0.07v \]
This equation shows that acceleration depends on the velocity of the boat and is in the opposite direction (negative sign). Solving for time with the adjusted velocity, we find:
\[ t = \frac{\ln(0.5)}{-0.07} \approx 9.9 \ s \]
This calculation ultimately shows the time required for the boat's speed to decrease to \(45 \ km/h\).

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Most popular questions from this chapter

A high-speed railway car goes around a flat, horizontal circle of radius \(470 \mathrm{~m}\) at a constant speed. The magnitudes of the horizontal and vertical components of the force of the car on a \(51.0 \mathrm{~kg}\) passenger are \(210 \mathrm{~N}\) and \(500 \mathrm{~N}\), respectively. (a) What is the magnitude of the net force (of all the forces) on the passenger? (b) What is the speed of the car?

Suppose you are sitting in a car that is speeding up. Assume the car has rear-wheel drive. (a) Draw free-body diagrams for your own body, the seat in which you are sitting (apart from the car), the car (apart from the seat), and the road surface where the tires and the road interact. (b) Describe each force in words; show larger forces with longer arrows. (c) Identify the third-law pairs of forces. (d) Explain carefully in your own words the origin of the force imparting acceleration to the car.

An old streetcar rounds a flat corner of radius \(9.1 \mathrm{~m}\), at \(16 \mathrm{~km} / \mathrm{h}\). What angle with the vertical will be made by the loosely hanging hand straps?

We know that as an object passes through the air, the air exerts a resistive force on it. Suppose we have a spherical object of radius \(R\) and mass \(m\). What might the force plausibly depend on? \- It might depend on the properties of the object. The only ones that seem relevant are \(m\) and \(R\). \- It might depend on the object's coordinate and its derivatives: \(\vec{r}, \vec{v}, \vec{a}, \ldots\) \- It might depend on the properties of the air, such as the density, \(\rho\). (a) Explain why it is plausible that the force the air exerts on a sphere depends on \(R\) but implausible that it depends on \(m\). (b) Explain why it is plausible that the force the air exerts depends on the object's speed through it, \(|\vec{v}|\), but not on its position, \(\vec{r}\), or acceleration, \(\vec{a}\). (c) Dimensional analysis is the use of units (e.g., meters, seconds, or newtons) associated with quantities to reason about the relationship between the quantities. Using dimensional analysis, construct a plausible form for the force that air exerts on a spherical body moving through it.

Sliding Stones Along the remote Racetrack Playa in Death Valley. California, stones sometimes gouge out prominent trails in the desert floor, as if they had been migrating (Fig 6-41). For years curiosity mounted about why the stones moved. One explanation was that strong winds during the occasional rainstorms would drag the rough stones over ground softened by rain. When the desert dried out, the trails behind the stones were hard-baked in place. According to measurements, the coefficient of kinetic friction between the stones and the wet playa ground is about \(0.80\). What horizontal force is needed on a stone of typical mass \(20 \mathrm{~kg}\) to maintain the stone's motion once a gust has started it moving? (Story continues with Problem 42.)
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