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Chapter 6: Problem 33

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 \(\vec{f}_{k}\) between boat and water is proportional to the speed \(v\) of the boat: \(f_{k}=70 v\), where \(v\) is in meters per second and \(f_{k}\) is in newtons. Find the time required for the boat to slow to \(45 \mathrm{~km} / \mathrm{h}\).

Short Answer

Expert verified
The boat takes approximately 9.9 seconds to slow to 45 km/h.

Step by step solution

01

Convert Units

First, we need to convert the boat's speed from kilometers per hour to meters per second. The initial speed is \(90\ \mathrm{km/h}\). To convert to \(\mathrm{m/s}\), use the conversion factor \(1\ \mathrm{km/h} = \frac{1}{3.6}\ \mathrm{m/s}\). Therefore, \(v_0 = 90 \times \frac{1}{3.6} = 25 \ \mathrm{m/s}\). Similarly, convert \(45\ \mathrm{km/h}\) to meters per second: \(v_f = 45 \times \frac{1}{3.6} = 12.5 \ \mathrm{m/s}\).
02

Express Force as a Differential Equation

The frictional force is given by \(f_k = 70v\). According to Newton鈥檚 second law, \(f = ma = m \frac{dv}{dt}\). Substituting for \(f_k\), we have \(70v = 1000 \frac{dv}{dt}\), which simplifies to \(\frac{dv}{dt} = \frac{70}{1000}v\).
03

Rearrange and Integrate the Differential Equation

Rearrange the differential equation to \(\frac{dv}{v} = -\frac{70}{1000} dt\) (the negative sign is added since the velocity decreases). Integrate both sides: the left for \(v\) and the right for \(t\). This yields \(\ln|v| = -\frac{70}{1000}t + C\).
04

Solve for the Constant of Integration

To find \(C\), use the initial condition \(v(0) = 25\ \mathrm{m/s}\). Substitute into the equation: \(\ln|25| = -\frac{70}{1000}\cdot0 + C\). Therefore, \(C = \ln 25\).
05

Solve for Time when Speed is 12.5 m/s

Substitute \(v = 12.5\ \mathrm{m/s}\) into the integrated equation: \(\ln|12.5| = -\frac{70}{1000} t + \ln 25\). Solving for \(t\), we get \(-\frac{70}{1000} t = \ln\frac{12.5}{25}\), which simplifies to \(t = \frac{1000}{70}\ln 2\). Calculate this to find \(t \approx 9.9\ \mathrm{seconds}\).

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Key Concepts

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

Differential Equations
In kinematics, differential equations are powerful tools used to model the behavior of dynamically changing systems like moving objects. In the context of the problem, we use a differential equation to model the rate of change of the boat's speed. Since the frictional force is proportional to the speed, we express this relationship mathematically as \[ f_k = 70v \] and invoke Newton鈥檚 Second Law \[ f = ma = m \frac{dv}{dt} \]. These allow us to write the differential equation \[ \frac{dv}{dt} = \frac{70}{1000}v \]. This equation captures how the speed of the boat changes over time due to the frictional force. By rearranging and integrating it, we can find how long it takes for the boat's speed to decrease to a certain level.
Newton's Second Law
Newton's Second Law is fundamental in understanding how forces affect motion. This law states that the force acting on an object is equal to its mass times its acceleration, expressed as \[ f = ma \]. In this context, the accelerating force is the frictional force exerted by water on the boat. Since the force is negative, it acts to decelerate the boat. The equation \[ 70v = 1000 \frac{dv}{dt} \] is derived by substituting the expression for the frictional force into Newton鈥檚 Second Law. This highlights that the boat's acceleration is directly related to its velocity, emphasizing how motion is influenced by external forces, like friction in this scenario.
Frictional Forces
Frictional forces play a critical role in motion mechanics, providing resistance that influences speed and movement. In this problem, the frictional force between the boat and water opposes the motion of the boat and slows it down. It's described by the equation \[ f_k = 70v \], indicating that the force increases linearly with speed. This relationship highlights a key property of friction: it depends on the velocity of the moving object. Consequently, as the boat slows, the force decreases, impacting how quickly the speed diminishes. Understanding the nature of this force is crucial in solving problems related to moving objects and is ubiquitous in everyday physics applications.
Unit Conversion
Unit conversion is an essential skill in physics to ensure consistency and accuracy in calculations. Here, we convert the boat's speeds from kilometers per hour to meters per second to make them compatible with the SI units used in physics calculations. The conversion is performed using the factor \[ 1 \text{ km/h} = \frac{1}{3.6} \text{ m/s} \]. This conversion is crucial as it ensures that all values are within the same unit system, preventing errors in further calculations. Understanding how and when to convert units is vital in scientific problem-solving, as it ensures clarity and consistency across different calculations and measurements.

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Most popular questions from this chapter

A locomotive accelerates a 25 -car train along a level track. Every car has a mass of \(5.0 \times 10^{4} \mathrm{~kg}\) and is subject to a frictional force \(f=250 v\), where the speed \(v\) is in meters per second and the force \(f\) is in newtons. At the instant when the speed of the train is \(30 \mathrm{~km} / \mathrm{h}\), the magnitude of its acceleration is \(0.20 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the tension in the coupling between the first car and the locomotive? (b) If this tension is equal to the maximum force the locomotive can exert on the train, what is the steepest grade up which the locomotive can pull the train at \(30 \mathrm{~km} / \mathrm{h}\) ?

An initially stationary box of sand is to be pulled across a floor by means of a cable in which the tension should not exceed \(1100 \mathrm{~N}\). The coefficient of static friction between the box and the floor is \(0.35\). (a) What should be the angle between the cable and the horizontal in order to pull the greatest possible amount of sand, and (b) what is the weight of the sand and box in that situation?

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?

\(.\Rightarrow F\) A circular-motion addict of mass \(80 \mathrm{~kg}\) rides a Ferris wheel around in a vertical circle of radius \(10 \mathrm{~m}\) at a constant speed of \(6.1 \mathrm{~m} / \mathrm{s}\). (a) What is the period of the motion? What is the magnitude of the normal force on the addict from the seat when both go through (b) the highest point of the circular path and (c) the lowest point?

You testify as an expert witness in a case involving an accident in which car \(A\) slid into the rear of car \(B\), which was stopped at a red light along a road headed down a hill (Fig. \(6-25\) ). You find that the slope of the hill is \(\theta=12.0^{\circ}\), that the cars were separated by distance \(d=24.0 \mathrm{~m}\) when the driver of car \(A\) put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car \(A\) at the onset of braking was \(v_{0}=18.0 \mathrm{~m} / \mathrm{s}\). With what speed did car \(A\) hit car \(B\) if the coefficient of kinetic friction was (a) \(0.60\) (dry road surface) and (b) \(0.10\) (road surface covered with wet leaves)?
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