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Chapter 11: Problem 17

The acceleration of a particle is defined by the relation \(a=-k / x\) It has been experimentally determined that \(v=15\) fts when \(x=0.6 \mathrm{ft}\) and that \(v=9 \mathrm{ft} / \mathrm{s}\) when \(x=1.2\) ft. Determine (a) the velocity of the particle when \(x=1.5 \mathrm{ft},(b)\) the position of the particle at which its velocity is zero.

Short Answer

Expert verified
Velocity at 1.5 ft is found by solving the equation with known constants. Find position for zero velocity by setting velocity equation to 0.

Step by step solution

01

Understand the Equation for Acceleration

The given acceleration of the particle is defined by the equation \(a = -\frac{k}{x}\). This means that acceleration is inversely proportional to position \(x\) and affected by a constant \(k\). To find \(k\), we need to relate the information from the velocity measurements.
02

Relate Acceleration to Velocity and Position

Recall that acceleration \(a\) is the derivative of velocity \(v\) with respect to time \(t\), and also that \(a = v \cdot \frac{dv}{dx}\) (since \(\frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}\)). Therefore, replace \(a\) with \(v \cdot \frac{dv}{dx}\) and substitute the given acceleration equation: \(-\frac{k}{x} = v \cdot \frac{dv}{dx}\).
03

Integrate to Find Velocity in Terms of Position

Separate variables via \(-k dx = v dv\) and integrate both sides to find the velocity equation. The left side becomes \(-k \ln|x| + C_1\) and the right side becomes \(\frac{v^2}{2} + C_2\), making use of integration basics. Solve for \(v^2\):\[v^2 = -2k \ln|x| + C\]
04

Determine the Constant Using Initial Conditions

To find the constant \(C\), substitute in one of the known conditions such as \(v=15\) fts when \(x=0.6\) ft: \[15^2 = -2k \ln(0.6) + C\].Do similarly for the second condition: \[9^2 = -2k \ln(1.2) + C\].By solving these two equations simultaneously, we can determine values for \(k\) and \(C\).
05

Solve for Velocity at Specific Position

Using the derived equation and the known constants \(k\) and \(C\), plug \(x=1.5\) ft into the equation \(v^2 = -2k \ln(x) + C\) to find the velocity at this position.
06

Determine Position When Velocity is Zero

Set the velocity \(v = 0\) and solve the equation \(0 = -2k \ln(x) + C\) for \(x\). This gives the position where the velocity is zero. Solve for \(x\) using logarithmic properties.

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Key Concepts

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

Inversely Proportional
When something is described as being inversely proportional, it means that as one value increases, the other decreases. In this exercise, the acceleration of the particle (\(a = -\frac{k}{x}\)) is inversely proportional to its position \(x\). This means that when the particle is closer to a point, its acceleration is greater.

To break it down, "inversely" suggests a reciprocal relationship. If the position \(x\) doubles, the acceleration halves, considering a constant \(k\). Understanding this inverse relationship is crucial when examining how forces affect motion in physics.

This principle is commonly observed not only in physics but in chemistry and economics as well, wherever balance between variables is crucial.
Velocity Measurement
Velocity is essentially the speed of a particle in a given direction. In this problem, velocities at specific positions are provided as conditions to determine further unknowns.

Here, the velocity measurements (\(v = 15\) ft/s at \(x = 0.6\) ft and \(v = 9\) ft/s at \(x = 1.2\) ft) serve as initial conditions to solve for the velocity equation.

The consistent direction of motion allows us to use these measurements to get valuable insights into the particle's dynamics and to calculate the constants in our motion equations. Without accurate velocity measurements or understanding their significance, predicting the motion becomes challenging.
Integral Calculus
Integral calculus deals with accumulation, such as areas under curves or total change over time. In this exercise, calculus is used specifically to determine velocity from acceleration.

We begin by expressing acceleration as a function of velocity and integrating, \[-k dx = v dv\]. This requires integrating both sides separately.

On integrating, the position side gives \(-k \ln|x| + C_1\) and the velocity side gives \(\frac{v^2}{2} + C_2\). These integrations help find the relation between velocity and position, providing us with the equation: \[v^2 = -2k \ln|x| + C\]. Understanding integral calculus is essential for students to bridge between acceleration and velocity in kinematics.
Initial Conditions
Initial conditions are given values of variables at a starting point often used to solve differential equations. They ground our mathematical models in reality.

In this problem, initial conditions are the given velocities at certain positions. For instance, \(v = 15\) ft/s at \(x = 0.6\) ft is one initial condition used to calculate unknown constants \(k\) and \(C\) in our integration result.

These conditions help solve equations meaningfully instead of abstractly. By plugging initial conditions into an equation, we solve real-world problems and adjust our functions to fit actual data.

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Most popular questions from this chapter

During a parassiling ride, the boat is traveling at a constant \(30 \mathrm{km} / \mathrm{hr}\) with a \(200-\mathrm{m}\) long tow line. At the instant shown, the angle between the line and the water is \(30^{\circ}\) and is increasing at a constant rate of \(2 \%\). Determine the velocity and acceleration of the parasailer at this instant.

Instruments in an airplane which is in level flight indicate that the velocity relative to the air (airspeed) is \(120 \mathrm{km} / \mathrm{h}\) and the direction of the relative velocity vector (heading) is \(70^{\circ}\) east of north. Instruments on the ground indicate that the velocity of the airplane (ground speed) is \(110 \mathrm{km} / \mathrm{h}\) and the direction of flight (course) is \(60^{\circ}\) east of north. Determine the wind speed and direction.

The motion of a particle is defined by the relation \(x=2 t^{3}-9 t^{2}+\) \(12 t+10,\) where \(x\) and \(t\) are expressed in feet and seconds, respectively. Determine the time, the position, and the acceleration of the particle when \(v=0\)

In a water-tank test involving the launching of a small model boat, the model's initial horizontal velocity is \(6 \mathrm{m} / \mathrm{s}\) and its horizontal acceleration varies linearly from \(-12 \mathrm{m} / \mathrm{s}^{2}\) at \(t=0\) to \(-2 \mathrm{m} / \mathrm{s}^{2}\) at \(t=t_{1}\) and then remains equal to \(-2 \mathrm{m} / \mathrm{s}^{2}\) until \(t=1.4 \mathrm{s}\). Knowing that \(v=1.8 \mathrm{m} / \mathrm{s}\) when \(t=t_{1}\), determine \((a)\) the value of \(t_{1},(b)\) the velocity and the position of the model at \(t=1.4 \mathrm{s}\).

The three-dimensional motion of a particle is defined by the position vector \(r=\left(R t \cos \omega_{n} t\right) \mathbf{i}+c t \mathbf{j}+\left(R t \sin \omega_{n} t\right) \mathbf{k} .\) Determine the magnitudes of the velocity and acceleration of the particle. (The space curve described by the particle is a conic helix.)
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