2.3E: Exercises for Section 2.3

Exercise \(\PageIndex{1}\)

A firefighter who weighs \(192\) lb slides down an infinitely long fire pole that exerts a frictional resistive force with magnitude proportional to his speed, with \(k=2.5\) lb-s/ft. Assuming that he starts from rest, find his velocity as a function of time and find his terminal velocity.

Answer

\( v = -\frac{384}{5}(1-e^{5t/12});  -\frac{384}{5}\text { ft/s} \)

Exercise \(\PageIndex{2}\)

A firefighter who weighs \(192\) lb slides down an infinitely long fire pole that exerts a frictional resistive force with magnitude proportional to her speed, with constant of proportionality \(k\). Find \(k\), given that her terminal velocity is \(-16\) ft/s, and then find her velocity \(v\) as a function of \(t\). Assume that she starts from rest.

Answer

\( k=12; v = -16(1 - e^{-2t}). \)

Exercise \(\PageIndex{3}\)

A boat weighs \(64,000\) lb. Its propellor produces a constant thrust of \(50,000\) lb and the water exerts a resistive force with magnitude proportional to the speed, with \(k=2000\) lb-s/ft. Assuming that the boat starts from rest, find its velocity as a function of time, and find its terminal velocity.

Answer

\( v = 25(1- e^{-t}); 25\text { ft/s} \)

Exercise \(\PageIndex{4}\)

A constant horizontal force of \(10\) N pushes a \(20\) kg-mass through a medium that resists its motion with \(.5\) N for every m/s of speed. The initial velocity of the mass is \(7\) m/s in the direction opposite to the direction of the applied force. Find the velocity of the mass for \(t > 0\).

Answer

\( v = 20 - 27 e^{-t/40}. \)

Exercise \(\PageIndex{5}\)

A stone weighing \(1/2\) lb is thrown upward from an initial height of \(5\) ft with an initial speed of \(32\) ft/s. Air resistance is proportional to speed, with \(k=1/128\) lb-s/ft. Find the maximum height attained by the stone.

Answer

\( \approx 17.10 \text { ft} \)

Exercise \(\PageIndex{6}\)

A \(3200\)-lb car is moving at \(64\) ft/s down a \(30\)-degree grade when it runs out of fuel. Find its velocity after that if friction exerts a resistive force with magnitude proportional to the square of the speed, with \(k=1\ \mbox{lb-s}^2/{\mbox ft}^2\). Also find its terminal velocity.

Answer

\( v = -\frac {40(13+3e^{-4t/5})}{13-3e^{-4t/5}}; -40 \text { ft/s} \)

Exercise \(\PageIndex{7}\)

A \(96\) lb weight is dropped from rest in a medium that exerts a resistive force with magnitude proportional to the speed. Find its velocity as a function of time if its terminal velocity is \(-128\) ft/s.

Answer

\( v = -128(1- e^{-t/4}) \)

Exercise \(\PageIndex{8}\)

An object with mass \(m\) moves vertically through a medium that exerts a resistive force with magnitude proportional to the speed. Let \(y=y(t)\) be the altitude of the object at time \(t\), with \(y(0)=y_0\). Use the results of Example 4.3.1 to show that

\[y(t)=y_0+{m\over k}(v_0-v-gt).\nonumber \]

Exercise \(\PageIndex{9}\)

An object with mass \(m\) is launched vertically upward with initial velocity \(v_0\) from Earth’s surface (\(y_0=0\)) in a medium that exerts a resistive force with magnitude proportional to the speed. Find the time \(T\) when the object attains its maximum altitude \(y_m\). Then use the result of Exercise 2.3.8 to find \(y_m\).

Answer

\(  T= \frac{m}{k}\ln (1+\frac {v_0k}{mg}); y_m=y_0=\frac{m}{k}\left[v_0- \frac{mg}{k}\ln (1+\frac {v_0k}{mg}) \right]  \)

Exercise \(\PageIndex{10}\)

An object weighing \(256\) lb is dropped from rest in a medium that exerts a resistive force with magnitude proportional to the square of the speed. The magnitude of the resisting force is \(1\) lb when \(|v|=4\ \mbox{ft/s}\). Find \(v\) for \(t > 0\), and find its terminal velocity.

Answer

\( v = \frac{64(1 - e^t)}{1 + e^t}, \) or \( v = -\frac{64(1 - e^{-t})}{1 + e^{-t}}. \) Therefore, \( \lim_{t \to \infty} v(t) = -64\text { ft/s}. \)

Exercise \(\PageIndex{11}\)

An object with mass \(m\) is given an initial velocity \(v_0\le0\) in a medium that exerts a resistive force with magnitude proportional to the square of the speed. Find the velocity of the object for \(t > 0\), and find its terminal velocity.

Answer

\(v = \alpha \frac{v_0(1+e^{-\beta t})-\alpha(1- e^{-\beta t})}{\alpha(1+e^{-\beta t})-v_0(1- e^{-\beta t})}; -\alpha \text { where}\) : \( \alpha= \sqrt \frac{mg}{k}, \beta= \sqrt \frac{kg}{m} \)

Exercise \(\PageIndex{12}\)

An object with mass \(m\) is launched vertically upward with initial velocity \(v_0\) in a medium that exerts a resistive force with magnitude proportional to the square of the speed.

1. Find the time \(T\) when the object reaches its maximum altitude.
2. Use the result of Exercise 2.3.11 to find the velocity of the object for \(t > T\).

Answer
 

1. \( T = \sqrt \frac{m}{kg} \tan^{-1} \left (v_0 \sqrt \frac{k}{mg}\right)v= -\sqrt \frac{mg}{k} \)
2. \( \frac{1 - e^{-2\sqrt{\frac{ak}{m}}(t-T)}}{1 + e^{-2\sqrt{\frac{ak}{m}}(t-T)}} \)

Exercise \(\PageIndex{13}\)

An object with mass \(m\) is given an initial velocity \(v_0\le0\) in a medium that exerts a resistive force of the form \(a|v|/(1+|v|)\), where \(a\) is positive constant.

1. Set up a differential equation for the speed of the object.
2. Use your favorite numerical method to solve the equation you found in (a), to convince yourself that there’s a unique number \(a_0\) such that \(\lim_{t\to\infty}s(t)=\infty\) if \(a\le a_0\) and \(\lim_{t\to\infty}s(t)\) exists (finite) if \(a>a_0\). (We say that \(a_0\) is the bifurcation value of \(a\). Try to find \(a_0\) and \(\lim_{t\to\infty}s(t)\) in the case where \(a>a_0\).)

Answer

\( s'= mg-\frac{as}{s+1}; a_0 = mg  \)

Exercise \(\PageIndex{14}\)

An object of mass \(m\) falls in a medium that exerts a resistive force \(f=f(s)\), where \(s=|v|\) is the speed of the object. Assume that \(f(0)=0\) and \(f\) is strictly increasing and differentiable on \((0,\infty)\).

1. Write a differential equation for the speed \(s=s(t)\) of the object. Take it as given that all solutions of this equation with \(s(0)\ge0\) are defined for all \(t>0\) (which makes good sense on physical grounds).
2. Show that if \(\lim_{s\to\infty}f(s)\le mg\) then \(\lim_{t\to\infty}s(t)=\infty\).
3. Show that if \(\lim_{s\to\infty}f(s)>mg\) then \(\lim_{t\to\infty}s(t)=s_T\) (terminal speed), where \(f(s_T)=mg\).

Answer

1. \( mv' = -mg + f(|v|) \); (A) \( ms' = mg - f(s). \)
2. Since \( f \) is increasing and \( \lim_{s \to \infty} f(s) \leq mg \), we have \( mg - f(s) > 0 \) for all \( s \). This and (A) imply that \( s \) is an increasing function of \( t \), so either \( \text{(B)} \quad \lim_{t \to \infty} s(t) = \infty, \) or \( \text{(C)} \quad \lim_{t \to \infty} s(t) = \bar{s} < \infty. \) However, (A) and (C) imply that \( s'(t) > K = \frac{g - f(\bar{s})}{m} \quad \text{for all} \quad t > 0. \) Consequently, \( s(t) > s_0 + Kt \quad \text{for all} \quad t > 0, \) which contradicts (C) because \( K > 0 \).
3. There is a unique positive number \( s_T \) such that \( f(s_T) = mg \), and \( s = s_T \) is a constant solution of (A). Now suppose that \( s(0) < s_T \). Then Theorem 1.7.1 implies that \( \text{(D)} \quad s(t) < s_T \quad \text{for all} \quad t > 0, \) so (A) implies that \( s \) is strictly increasing. This and (D) imply that \( \lim_{t \to \infty} s(t) = \bar{s} \leq s_T. \) If \( \bar{s} < s_T \), then (A) implies that \( s'(t) > K = \frac{g - f(\bar{s})}{m}. \) Consequently, \( s(t) > s(0) + Kt, \) which contradicts (D) because \( K > 0 \). Therefore, \( s(0) < s_T \Rightarrow \lim_{t \to \infty} s(t) = s_T. \) A similar proof with inequalities reversed shows that \( s(0) > s_T \Rightarrow \lim_{t \to \infty} s(t) = s_T. \)

Exercise \(\PageIndex{15}\)

A \(100\)-g mass with initial velocity \(v_0\le0\) falls in a medium that exerts a resistive force proportional to the fourth power of the speed. The resistance is \(.1\) N if the speed is \(3\) m/s.

1. Set up the initial value problem for the velocity \(v\) of the mass for \(t>0\).
2. Use Exercise 2.3E.14 part (c) to determine the terminal velocity of the object.

Answer

1. \( v' = -9.8+\frac{v^4}{81} \)
2. \( v_T \approx -5.308\text { m/s} \)

Exercise \(\PageIndex{16}\)

A \(64\)-lb object with initial velocity \(v_0\le0\) falls through a dense fluid that exerts a resistive force proportional to the square root of the speed. The resistance is \(64\) lb if the speed is \(16\) ft/s.

1. Set up the initial value problem for the velocity \(v\) of the mass for \(t>0\).
2. Use Exercise 2.3E.14 part (c) to determine the terminal velocity of the object.

Answer

1. \( v' = -32 + 8 \sqrt{|v|}. \)
2. From Exercise 2.3E.14 part (c), \( v_T \) is the negative number such that \( -32 + 8 \sqrt{|v_T|} = 0; \) thus, \( v_T = -16 \text{ ft/s}. \)

Exercise \(\PageIndex{17}\)

A space probe is to be launched from a space station \(200\) miles above Earth. Determine its escape velocity in miles/s. Take Earth’s radius to be \(3960\) miles.

Answer

\( \approx 6.76 \text{ miles/s}  \)

Exercise \(\PageIndex{18}\)

A space vehicle is to be launched from the moon, which has a radius of about \(1080\) miles. The acceleration due to gravity at the surface of the moon is about \(5.31\) ft/s\(^2\). Find the escape velocity in miles/s.

Answer

\( v_e = \sqrt{\frac{2 \cdot 5.31 \cdot 1080}{5280}} \approx 1.47 \text{ miles/s}. \)

Exercise \(\PageIndex{19}\)

1. Show that Equation (2.3.27) can be rewritten as \[v^2={h-y\over y+R} v^2_e+v_0^2.\nonumber \]
2. Show that if \(v_0=\rho v_e\) with \(0\le \rho < 1\), then the maximum altitude \(y_m\) attained by the space vehicle is \[y_m={h+R\rho^2\over 1-\rho^2}.\nonumber \]
3. By requiring that \(v(y_m)=0\), use Equation (2.3.26) to deduce that if \(v_0 < v_e\) then \[|v|=v_e\left[{(1-\rho^2)(y_m-y)\over y+R}\right]^{1/2},\nonumber \] where \(y_m\) and \(\rho\) are as defined in (b) and \(y \ge h\).
4. Deduce from (c) that if \(v < v_e\), the vehicle takes equal times to climb from \(y=h\) to \(y=y_m\) and to fall back from \(y=y_m\) to \(y=h\).

Exercise \(\PageIndex{20}\)

In the situation considered in the discussion of escape velocity, show that \(\lim_{t\to\infty}y(t)=\infty\) if \(v(t)>0\) for all \(t>0\).

Answer

\( \alpha = \frac{gR^2}{(y_m+R)^2}  \)