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A \(1400-\mathrm{kg}\) automobile starts from rest and travels \(400 \mathrm{m}\) during a performance test. The motion of the automobile is defined by the relation \(x=4000 \mathrm{ln}(\cosh 0.03 t),\) where \(x\) and \(t\) are expressed in meters and seconds, respectively. The magnitude of the aerodynamic drag is \(D=0.35 v^{2},\) where \(D\) and \(v\) are expressed in newtons and \(\mathrm{m} / \mathrm{s},\) respectively. Determine the power dissipated by the aerodynamic drag when \((a) t=10 \mathrm{s},(b) t=15 \mathrm{s}\)

Short Answer

Expert verified
The power dissipated at (a) 10 s is approximately 127.9 W and at (b) 15 s is approximately 363.0 W.

Step by step solution

01

Find the velocity equation

The relation for motion is given as \(x=4000 \ln(\cosh(0.03t))\). To find velocity \(v(t)\), we need the derivative of \(x(t)\) with respect to time \(t\). \[v(t) = \frac{dx}{dt} = 4000 \times \frac{d}{dt}(\ln(\cosh(0.03t)))\]This yields: \[v(t) = 4000 \times 0.03 \times \tanh(0.03t)\]\[v(t) = 120 \times \tanh(0.03t)\]
02

Find the velocity at specific times

We calculate the velocity at \(t = 10\) s and \(t = 15\) s using the velocity function found in Step 1. For \(t = 10\) s, \[v(10) = 120 \times \tanh(0.3)\].For \(t = 15\) s, \[v(15) = 120 \times \tanh(0.45)\].
03

Calculate aerodynamic drag force

Use the given formula for aerodynamic drag \(D = 0.35 v^2\).At \(t = 10\) s, substitute \(v(10)\) into the formula.\[D = 0.35 \times (v(10))^2\].At \(t = 15\) s, substitute \(v(15)\) into the formula.\[D = 0.35 \times (v(15))^2\].
04

Calculate power dissipated by aerodynamic drag

The power dissipated by aerodynamic drag is calculated using:\[P = D \times v\].For \(t = 10\) s, calculate \(P = D(10) \times v(10)\).For \(t = 15\) s, calculate \(P = D(15) \times v(15)\).
05

Substitute and solve

For \(t=10\) s, substitute the calculated values of \(v(t)\) and \(D\) into the power formula and solve.For \(t=15\) s, repeat these calculations with the corresponding velocity and drag force values to solve for the power.

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Key Concepts

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

Automobile Dynamics
Automobile dynamics is the study of the forces and movements that affect vehicles. Understanding these dynamics is crucial for designing cars that are safe and perform well. When we look at a car driving in a straight line, several forces are at play, including the engine force, frictional forces, and aerodynamic forces.

In vehicle dynamics, the car's acceleration and velocity are determined by these forces and its mass. For a car starting from rest, as in the exercise, acceleration builds as it travels over a distance. The car's dynamics change from simply overcoming inertia to dealing with other forces like aerodynamic drag as speed increases.

Analyzing automobile dynamics involves understanding how speed changes with time, the impact of external forces, and how these factors contribute to the vehicle's overall performance.
Aerodynamic Drag
Aerodynamic drag is a force that opposes the forward motion of a vehicle. It arises due to air resistance as the car moves through the atmosphere. The magnitude of this force often depends on the velocity of the vehicle.

In the provided exercise, the aerodynamic drag is expressed through the formula \(D = 0.35 v^2\). This indicates that drag increases with the square of the velocity. This relationship highlights why higher speeds lead to significantly greater energy requirements for maintaining or increasing speed, as drag becomes a dominant force.

Understanding aerodynamic drag is essential for improving vehicle efficiency. By reducing drag through design optimizations, vehicles can achieve better fuel economy or stronger performance at high speeds.
Power Dissipation
Power dissipation in the context of vehicles refers to the power that is lost, often through forces like aerodynamic drag and friction. In this problem, it specifically involves the power expended to overcome aerodynamic drag.

The power dissipated by aerodynamic drag is calculated using the formula \(P = D \times v\), where \(D\) is the drag force and \(v\) is the velocity. This relation highlights how power requirements increase with both the force of the drag and the speed of the vehicle.

For efficient vehicle design, minimizing power dissipation is crucial as it leads to better fuel efficiency and performance. This concept underscores why engineering efforts often focus on reducing drag through aerodynamics.
Velocity Calculation
Velocity calculation is crucial in understanding how a vehicle moves over time. In dynamic systems, velocity represents the rate of change of displacement with time.

In the given exercise, the velocity function was derived from the position equation given by \(x = 4000 \ln(\cosh(0.03t))\). Calculating the derivative of this equation with respect to time provides the velocity formula: \(v(t) = 120 \times \tanh(0.03t)\).

This velocity function allows us to determine how fast the car is moving at any given time \(t\). For instance, specific calculations at \(t = 10\) seconds and \(t = 15\) seconds provide specific velocities, which can then be used to explore other physics concepts such as drag force and power dissipation.

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Most popular questions from this chapter

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