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Chapter 5: Q34E (page 273)

Falling Object. The motion of an object moving vertically through the air is governed by the equation\(\frac{{{{\bf{d}}^{\bf{2}}}{\bf{y}}}}{{{\bf{d}}{{\bf{t}}^{\bf{2}}}}}{\bf{ = - g - }}\frac{{\bf{g}}}{{{{\bf{V}}^{\bf{2}}}}}\frac{{{\bf{dy}}}}{{{\bf{dt}}}}\left| {\frac{{{\bf{dy}}}}{{{\bf{dt}}}}} \right|\)where y is the upward vertical displacement and V is a constant called the terminal speed. Take \({\bf{g = 32ft/se}}{{\bf{c}}^{\bf{2}}}\)and V = 50 ft/sec. Sketch trajectories in the yv-phase plane for \( - 100 \le {\bf{y}} \le 100, - 100 \le {\bf{v}} \le 100\)starting from y = 0 and y = -75, -50, -25, 0, 25, 50, and 75 ft/sec. Interpret the trajectories physically; why is V called the terminal speed?

Short Answer

Expert verified

The terminal velocity is the maximum velocity attainable by an object as it falls through the air.

Step by step solution

01

Find trajectory

The trajectory approaches the value\(v = - 50\). This means that the object will be falling

through the air at\(50{\rm{ }}ft/sec\). This is the terminal velocity and that velocity is the maximum

velocity attainable by the object as it falls through the air.

02

Step 2:The trajectory

The trajectory can be plotted by mat lab.

Therefore, the terminal velocity is the maximum velocity attainable by an object as it falls through the air.

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Most popular questions from this chapter

Suppose the displacement functions xtandyt for a coupled mass-spring system (similar to the one discussed in Problem 6) satisfy the initial value problem

x''(t)+5x(t)-2y(t)=0y''(t)+2y(t)-2x(t)=3sin2tx(0)=x'(0)=0y(0)=1,y'(0)=0

Solve forxt andyt

In Section 3.6, we discussed the improved Euler鈥檚 method for approximating the solution to a first-order equation. Extend this method to normal systems and give the recursive formulas for solving the initial value problem.

In Problems 11鈥�14, solve the related phase plane differential equation for the given system. Then sketch by hand several representative trajectories (with their flow arrows).

dxdt=-8y,dydt=18x

A Problem of Current Interest. The motion of an ironbar attracted by the magnetic field produced by a parallel current wire and restrained by springs (see Figure 5.17) is governed by the equation\(\frac{{{{\bf{d}}^{\bf{2}}}{\bf{x}}}}{{{\bf{d}}{{\bf{t}}^{\bf{2}}}}}{\bf{ = - x + }}\frac{{\bf{1}}}{{{\bf{\lambda - x}}}}\) for \({\bf{ - }}{{\bf{x}}_{\bf{o}}}{\bf{ < x < \lambda }}\)where the constants \({{\bf{x}}_{\bf{o}}}\) and \({\bf{\lambda }}\) are, respectively, the distances from the bar to the wall and to the wire when thebar is at equilibrium (rest) with the current off.

  1. Setting\({\bf{v = }}\frac{{{\bf{dx}}}}{{{\bf{dt}}}}\), convert the second-order equation to an equivalent first-order system.
  2. Solve the related phase plane differential equation for the system in part (a) and thereby show that its solutions are given by\({\bf{v = \pm }}\sqrt {{\bf{C - }}{{\bf{x}}^{\bf{2}}}{\bf{ - 2ln(\lambda - x)}}} \), where C is a constant.
  3. Show that if \({\bf{\lambda < 2}}\) there are no critical points in the xy-phase plane, whereas if \({\bf{\lambda > 2}}\) there are two critical points. For the latter case, determine these critical points.
  4. Physically, the case \({\bf{\lambda < 2}}\)corresponds to a current so high that the magnetic attraction completely overpowers the spring. To gain insight into this, use software to plot the phase plane diagrams for the system when \({\bf{\lambda = 1}}\) and when\({\bf{\lambda = 3}}\).
  5. From your phase plane diagrams in part (d), describe the possible motions of the bar when \({\bf{\lambda = 1}}\) and when\({\bf{\lambda = 3}}\), under various initial conditions.

Show that the Poincare map for equation (1) is not chaoticby showing that if\({\bf{(}}{{\bf{x}}_{\bf{o}}}{\bf{,}}{{\bf{\nu }}_{\bf{o}}}{\bf{)}}\)and\({\bf{(}}{{\bf{x}}^{\bf{*}}}_{\bf{o}}{\bf{,}}{{\bf{\nu }}^{\bf{*}}}_{\bf{o}}{\bf{)}}\)are two initial values that define the Poincare maps\({\bf{(}}{{\bf{x}}_{\bf{n}}}{\bf{,}}{{\bf{\nu }}_{\bf{n}}}{\bf{)}}\) and\({\bf{(}}{{\bf{x}}^{\bf{*}}}_{\bf{n}}{\bf{,}}{{\bf{\nu }}^{\bf{*}}}_{\bf{n}}{\bf{)}}\), respectively, using the recursive formulas in (3), then one can make the distance between\({\bf{(}}{{\bf{x}}_{\bf{n}}}{\bf{,}}{{\bf{\nu }}_{\bf{n}}}{\bf{)}}\)and\({\bf{(}}{{\bf{x}}^{\bf{*}}}_{\bf{n}}{\bf{,}}{{\bf{\nu }}^{\bf{*}}}_{\bf{n}}{\bf{)}}\)small by making the distance between\({\bf{(}}{{\bf{x}}_{\bf{o}}}{\bf{,}}{{\bf{\nu }}_{\bf{o}}}{\bf{)}}\) and \({\bf{(}}{{\bf{x}}^{\bf{*}}}_{\bf{o}}{\bf{,}}{{\bf{\nu }}^{\bf{*}}}_{\bf{o}}{\bf{)}}\)small. (Hint: Let \({\bf{(A,}}\phi {\bf{)}}\)and \({\bf{(}}{{\bf{A}}^{\bf{*}}}{\bf{,}}{\phi ^ * }{\bf{)}}\) be the polar coordinates of two points in the plane. From the law of cosines, it follows that the distance d between them is given by\({{\bf{d}}^{\bf{2}}}{\bf{ = (A - }}{{\bf{A}}^{\bf{*}}}{{\bf{)}}^{\bf{2}}}{\bf{ + 2A}}{{\bf{A}}^{\bf{*}}}{\bf{(1 - cos(}}\phi {\bf{ - }}{\phi ^ * }{\bf{))}}\).)
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