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Chapter 3: Q 3.4-1E (page 115)

An object of mass 5 kg is released from rest 1000 m above the ground and allowed to fall under the influence of gravity. Assuming the force due to air resistance is proportional to the velocity of the object with proportionality constant b = 50 N-sec/m, determine the equation of motion of the object. When will the object strike the ground?

Short Answer

Expert verified
  • The equation of motion of the object isx(t)=0.981t-0.0981(1-e-10t)m
  • The time taken by the object is1019 sec.

Step by step solution

01

Find the equation of motion of an object

The given values are m = 5, g = 9.81,v0= 0, b = 50,

The equation of motion isrole="math" localid="1664170601861" x(t)=mgtb+mbv-mgb1-e-btm

Put all the given values

5(9.81)t50+550(0-5(9.81)50)(1-e-50t5)(9.81)t10-(9.81)100)(1-e-10t)x(t)=(9.81)t10-(9.81)100)(1-e-10t)x(t)=0.981t-0.0981(1-e-10t)m

Hence the equation of motion is role="math" localid="1664170624317" x(t)=0.981t-0.0981(1-e-10t)m

02

Finding when will the object strike the ground

1000=0.981t-0.0981(1-e-10t)1000=0.981t-0.0981(Becausevalueofissosmallsoneglecting)1000+0.0981=0.981t1000.0981=0.981tt=1000.0981/0.981t=1019sec

Hence, the object will strike the ground in 1019 sec.

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Local versus Global Error. In deriving formula (4) for Euler鈥檚 method, a rectangle was used to approximate the area under a curve (see Figure 3.14). With

\({\bf{g(t) = f(t,f(t))}}\), this approximation can be written as \(\int\limits_{{{\bf{x}}_{\bf{n}}}}^{{{\bf{x}}_{{\bf{n + 1}}}}} {{\bf{g(t)dt}} \approx {\bf{hg(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} \)where \({\bf{h = }}{{\bf{x}}_{{\bf{n + 1}}}}{\bf{ - }}{{\bf{x}}_{\bf{n}}}\) .

  1. Show that if g has a continuous derivative that is bounded in absolute value by B, then the rectangle approximation has error\(\left( {\bf{O}} \right){{\bf{h}}^{\bf{2}}}\); that is, for some constant M, \(\left| {\int\limits_{{{\bf{x}}_{\bf{n}}}}^{{{\bf{x}}_{{\bf{n + 1}}}}} {{\bf{g(t)dt - hg(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} } \right| \le {\bf{M}}{{\bf{h}}^{\bf{2}}}\).This is called the local truncation error of the scheme. [Hint: Write \(\int\limits_{{{\bf{x}}_{\bf{n}}}}^{{{\bf{x}}_{{\bf{n + 1}}}}} {{\bf{g(t)dt - hg(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} {\bf{ = }}\int\limits_{{{\bf{x}}_{\bf{n}}}}^{{{\bf{x}}_{{\bf{n + 1}}}}} {\left[ {{\bf{g(t)dt - g(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} \right]{\bf{dt}}} \). Next, using the mean value theorem, show that\(\left| {{\bf{g(t)dt - g(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} \right| \le {\bf{B}}\left| {{\bf{t - }}{{\bf{x}}_{\bf{n}}}} \right|\) . Then integrate to obtain the error bound\(\left( {\frac{{\bf{B}}}{{\bf{2}}}} \right){{\bf{h}}^{\bf{2}}}\).]
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