Solving for the distance

3 0 60-0 = 60 Paula traveled 3 hours at 0 mph. Doesn t make sense. 

Part III Tackling Word Problems from Algebra 

Several of the entries in the table are reasonable scenarios for Paula and Ken. To actually finish the problem, you need a little more information. Assume, for instance, that Ken drove 60 mph. That means that the bus Paula is riding on is driving at 40 mph. The equation now becomes 40f = 60(f- 3), which multiplies out to be 40f = 60f- 180. Simplifying, you get 180 = 20f or t = 9. If Paula rode 9 hours at 40 mph, then the distance from school to home is 40 x 9 = 

The Problem The contestant from Kenya can run the 10,000-meter race at an average of 6 meters per second. The contestant from Ethiopia has a best time so far of 5.5 meters per second. How far back should the runner from Kenya start to have the expectation that they would cross the finish line at the same time if they both start at the same time  

In this problem, the distances aren t really equal. You have to add on x number of meters to represent the additional distance that the runner from Kenya must cover. The rates are also different, but the times will be the same, if they start at the same time and finish at the same time. Take the distance formula, d = rt and solve for t, giving you t = —. Now set the time it takes the runner from Kenya equal to the time for the runner from Ethiopia. Then change the times to distances and rates. 

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With distance problems, you can solve for the distance traveled or the speed at which objects are traveling or the amount of time spent. The two problems in this section involve solving for how much time it takes to reach a goal. 

One might be tempted to derive the diffusion coefficient from the slope of the linear estimation in the penetration-distance plot if the relative concentration (C/Co) is known. Unfortunately, this is not trivial because eq 7, containing the error function, cannot be solved for the distance x to obtain a linear equation, the slope of which can then be solved for D. 

In these equations the independent variable x is the distance normal to the disk surface. The dependent variables are the velocities, the temperature T, and the species mass fractions Tit. The axial velocity is u, and the radial and circumferential velocities are scaled by the radius as F = vjr and W = wjr. The viscosity and thermal conductivity are given by /x and A. The chemical production rate cOjt is presumed to result from a system of elementary chemical reactions that proceed according to the law of mass action, and Kg is the number of gas-phase species. Equation (10) is not solved for the carrier gas mass fraction, which is determined by ensuring that the mass fractions sum to one. An Arrhenius rate expression is presumed for each of the elementary reaction steps. 

It is clear that once we select rectangular coordinates and eliminate convective transport we are left with the temperature profile predicted by a simple film model (discussed in Section III.A). By solving for the heat flux in one dimension it can be seen that the temperature profile is a simple linear function of distance, chamber temperature, and surface temperature. 

The main macrokinetic problem to be solved for the description of this reaction is finding the evolution of the profile of concentrations Mi, M2 of monomeric units Mi, M2 inside a globule with radius R. By virtue of the spherical symmetry of the problem, concentration M is the same at all points of a globule located at identical distance r from its center. The same condition is apparently met by the concentrations of the second type units M2 = Mj0 - Mi and low-molecular reagent Z. Presuming monomeric units to... 

Even if the initial value of the level-set function (x,0) is set to be the distance function, the level set function r/j may not remain as a distance function at t >() when the advection equation, Eq. (3), is solved for . Thus, a redistance scheme is needed to enforce the condition of V0 = 1. An iterative procedure was designed (Sussman et al., 1998) to reinitialize the level-set function at each time step so that the level-set function remains as a distance function while maintaining the zero level set of the level-set function. This is achieved by solving for the steady-state solution of the equation (Sussman et al., 1994, 1998 Sussman and Fatemi, 1999) ... 

Because of their importance as basic primary centers, we will now discuss the optical bands associated with the F centers in alkali halide crystals. The simplest approximation is to consider the F center - that is, an electron trapped in a vacancy (see Figure 6.12) - as an electron confined inside a rigid cubic box of dimension 2a, where a is the anion-cation distance (the Cr -Na+ distance in NaCl). Solving for the energy levels of such an electron is a common problem in quantum mechanics. The energy levels are given by... 

The mass transfer coefficient is usually much lower in the Hquid phase, and therefore is a function of R, the distance from the wall to the interface. One would have to solve for the steady-state profile C iR), and find its average CX(z) to insert into the PFTR mass-balance equations simultaneously to find Ca(L) in each phase. 

Person 1 Use Eq. (1.16) with the values of m and n of the Lennard-Jones potential to solve for the constant a in terms of b and the equilibrium bond distance, ro. Now perform the determination of F ax as given by Eq. (1.15) substitute this value of a back into Eq. (1.12), differentiate it twice with respect to r (remember that ro is a constant), and set this equal to zero (determine, then maximize the force function). Solve this equation for r in terms of tq. The other constant should drop out. 

We solved for the equilibrium bond distance, ro, in Eq. (1.16), and the constants a and b have, in effect, just been evaluated. Inserting these values into Eq. (1.25), along with... 

This problem has Shelly traveling one less than twice as far as Shirley. You could make a table of possible values for the distances they traveled. You d probably be interested only in whole-number values, which won t solve the problem if the answer is a fraction, but you may get fairly close to the answer. Table 4-1 has some possible numbers or distances, starting with Shirley going 1 mile and ending to keep the total distance from getting larger than 17. 

You re looking for the height of the prism — the distance between the two trapezoidal bases on either end of the trough. Find the area of the trapezoidal base (end) and insert it into the formula for the volume. Replace the V in the formula with 60 and solve for the height of the prism. 

Disks fall into the second category of electrode, at which nonlinear diffusion occurs. The lines of flux to a disk electrode (Figure 12.2B) do not coincide with the simple geometries for which we derived Fick s second law, and the diffusion problem must therefore be expressed in two dimensions. Note that a line passing through the center of the disk and normal to the plane of the disk is a cylindrical axis of symmetry, so it is sensible to choose the radial distance from this axis as one of the coordinates for the problem. Diffusion along this radial coordinate, r, is described by Equation 12.7. Also, diffusion along the coordinate, x, normal to the plane of the electrode is described by Equation 12.4. Thus, the form of Fick s second law that must be solved for the disk is ... 

The preceding analysis views the problem of solving for the sine-wave amplitudes and phases in the frequency domain. Alternatively, the problem can be viewed in the time domain. It has been shown that [Quatieri and Danisewicz, 1990], for suitable window lengths, the vectors a andJ3 that satisfy Equation (9.75) also approximate the vectors that minimize the weighted mean square distance between the speech frame and the steady state sinusoidal model for summed vocalic speech with the sinusoidal frequency vector . Specifically, the following minimization is performed with respect to a andJ3... 

For this non-isothermal flow consider a Newtonian fluid between two parallel plates separated by a distance h. Again we consider the notation presented in Fig. 6.58, however, with both upper and lower plates being fixed. We choose the same exponential viscosity model used in the previous section. We are to solve for the velocity profile between the two plates with an imposed pressure gradient in the x-direction and a temperature gradient in the y-direction. 

Where l is the length of the capillary, a is the radius of the capillary, Jo, is a Bessel function of the first kind and k = sj / r /poS) = 8, is the viscous skin depth, which is the distance at which the amplitude of the vorti-city (transverse) wave has attenuated by a factor of the natural logarithm e . Inserting (3) into (1) and using Poisson s equation for the charges distribution we can solve for the FDSP Helmholtz-Smoluchowski equation [Reppert et al., 2001],... 

Because the z values from Eqn. (7-30) will always lead to the same concentration (i.e. 34 mg/ kg), one can simply solve z for the distance ... 

In the above expressions, the subscripts p and w denote the reflections from the particle and the boundary wall, respectively. The superscript (z ) represents the z th reflection from the particle surface or the solid boundary. Note that Eqs. (58) and (59) are actually power series in k, which is the ratio of the particle radius a to the distance between the particle center and the boundary d. The advantage of this method is that one needs to consider boundary conditions associated with only one surface at a time. After solving for the electric field and the fluid velocity for each reflection, the electrophoretic velocity can be calculated and expressed by... 

We will come to the form of the interaction potential employed in the next section given the appropriate form, the coupled equations (10.165) were solved by a numerical method due to Johnson [218] and described by Hutson [217]. For most states the coupled equations were solved for intemuclear distances up to 25 A, and the energies of the levels converged to 2 MHz. A computer program for interactive nonlinear least squares fitting of the parameters to the observed transition frequencies has been described by Law and Hutson [219]. 

The method of characteristics, the distance method of lines (continuous-time discrete-space), and the time method of lines (continuous-space discrete-time) were used to solve the solids stream partial differential equations. Numerical stiffness was not considered a problem for the method of characteristics and time method of lines calculations. For the distance method of lines, a possible numerical stiffness problem was solved by using a simple sifting procedure. A variable-step fifth-order Runge-Kutta-Fehlberg method was used to integrate the differential equations for both the solids and the gas streams. 

This improvement was suggested by Heiller and London. If we solve for the energy assodated with Eq. 5.2. we obtain curve b in Fig 5.1. The energy has improved greatly (—303kJ mn ) and also the distance has improved slightly. Since... 

We now insert rate laws written in terms of molar flow rates [e.g., Equation (3-45)] into the mole balances (Table 6-1). After performing this operation for each species we arrive at a coupled set of first-order ordinary differential equations to be solved for the molar flow rates as a function of reactor volume (i.e., distance along the length of the reactor). In liquid-phase reactions, incorporating and solving for total molar flow rate is not necessary at each step along the solution pathway because there is no volume change with reaction. 

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