Question

An object of mass is moving horizontally through a medium which resists the motion with a force that is a function of the velocity; that is,

\(m\frac{{{d^2}s}}{{d{t^2}}} = m\frac{{dv}}{{dt}} = f(v)\)

where\(v = v(t)\) and \(s = s(t)\)represent the velocity and position of the object at time t, respectively. For example, think of a boat moving through the water.

(a) Suppose that the resisting force is proportional to the velocity, that is,\(f(v) = - kv\), k a positive constant. (This model is appropriate for small values of v.) Let\(v(0) = {v_0}\) and\(s(0) = {s_0}\) be the initial values of v and s. Determine v and s at any time t. What is the total distance that the object travels from time\(t = 0\) ?

(b) For larger values of v a better model is obtained by supposing that the resisting force is proportional to the square of the velocity, that is,\(f(v) = - k{v^2},k > 0\). (This model was first proposed by Newton.) Let\({v_0}\)and\({s_0}\)be the initial values of v and s. Determine v and s at any time t. What is the total distance that the object travels in this case?

Short answer

a)

The velocity is \(v(t) = {v_0}\).

The position is \(s(t) = \frac{{m{v_0}}}{k}\left( {1 - {e^{\frac{{ - kt}}{m}}}} \right) + {s_0}\).

The total distance travelled \(\frac{{m{v_0}}}{k} + {s_0}\).

b)

The velocity is \(v(t) = \left( {\frac{{m{v_0}}}{{{v_0}kt + m}}} \right)\).

The position is \(s(t) = \left( {\frac{m}{k}} \right)\ln \left( {\left| {\frac{{m + {v_0}kt}}{m}} \right|} \right) + {s_0}\).

The total distance travelled is infinite.

Step-by-step solution

Concept used

Velocity as a function of position definition:
Velocity is the rate of change of position. The change of position is the displacement which is the shortest distance between the initial position and the final position in a particular direction of an object.

Given Information from question

a)

The resisting force is \(f(v) = - kv\).

The given function is

\(m\frac{{{d^2}s}}{{d{t^2}}} = m\frac{{dv}}{{dt}} = f(v)\)

Calculate the velocity from given function

Given function is

\(m\frac{{dv}}{{dt}} = - kv\)

Separate the variables, \(\frac{{dv}}{v} = - \frac{k}{m}dt\)

Integrate on both sides

\(\int {\frac{{dv}}{v}} = - \int {\frac{k}{m}} dt\)

\(\ln |v| = - \frac{k}{m}t + C\) ……. (1)

Given initial condition is \(v(0) = {v_0}\)

\(\ln \left| {{v_0}} \right| = - \frac{k}{m}(0) + {C_1}\)

\({C_1} = \ln \left| {{v_0}} \right|\)

Substitute \({C_1}\) value in equation (1)

\(\ln |v| = - \frac{k}{m}t + \ln \left| {{v_0}} \right|\)

Take exponentials on both sides

\(v = {e^{ - \frac{k}{m}t + \ln \left| {{v_0}} \right|}}\)

\(v(t) = {e^{ - \frac{k}{m}t}} \cdot {v_0}\) ……. (2)

Therefore, the velocity is \(v(t) = {e^{ - \frac{k}{m}t}} \cdot {v_0}\).

The velocity for given time \(t = 0\)

\(v(t) = {v_0}\)

Calculate the position 

The function is

\(\frac{{ds}}{{dt}} = {e^{\frac{{ - kt}}{m}}} \cdot {v_0}\)

Integrate on both sides

\(\int d s = \int {{e^{\frac{{ - kt}}{m}}}} \cdot {v_0}s(t)\)

\( = - \frac{{m{v_0}}}{k}{e^{\frac{{ - kt}}{m}}} + {C_2}\) ……. (3)

Substitute the initial condition \(s(0) = {s_0}\) in equation (3),

\(\begin{aligned}{s_0} &= - \frac{{m{v_0}}}{k}{e^0} + {C_2}\\{s_0} + \frac{{m{v_0}}}{k} &= {C_2}\end{aligned}\)

Substitute \({C_2}\) value in equation (2)

\(\begin{aligned}s(t) &= - \frac{{m{v_0}}}{k}{e^{\frac{{ - kt}}{m}}} + {s_0} + \frac{{m{v_0}}}{k}s(t)\\ &= \frac{{m{v_0}}}{k}\left( {1 - {e^{\frac{{ - kt}}{m}}}} \right) + {s_0}\end{aligned}\)

Therefore, the position is \(s(t) = \frac{{m{v_0}}}{k}\left( {1 - {e^{\frac{{ - tk}}{m}}}} \right) + {s_0}\).

Calculate the distance travelled

The distance travelled time 0 to \(t = \) final position - initial stage.

The distance travelled \( = \mathop {\lim }\limits_{t \to \infty } \left( {s(t) - {s_0}} \right)\)

\(\mathop {\lim }\limits_{t \to \infty } \left( {s(t) - {s_0}} \right) = \mathop {\lim }\limits_{t \to \infty } \left( {\frac{{m{v_0}}}{k}\left( {1 - {e^{\frac{{ - kt}}{m}}}} \right) + {s_0}} \right) - {s_0}\)

\( = \mathop {\lim }\limits_{t \to \infty } \frac{{m{v_0}}}{k}\left( {1 - {e^{\frac{{ - kt}}{m}}}} \right)\)

\( = \frac{{m{v_0}}}{k}\)

The total distance travelled = initial stage + final position.

Therefore, the total distance travelled \(\frac{{m{v_0}}}{k} + {s_0}\).

Given information from question

b)

The function is \(m\frac{{{d^2}s}}{{d{t^2}}} = m\frac{{dv}}{{dt}} = f(v)\).

Here \(v(0) = {v_0}\) and \(s(0) = {s_0}\) be the initial value of \(v\) and \(s\).

Calculate the velocity by given function

The given function is

\(m\frac{{dv}}{{dt}} = - k{v^2}\)

Separate the variables,

\(\frac{{dv}}{{{v^2}}} = - \frac{k}{m}dt\)

Integrating on both sides,

\(\begin{aligned}\int {\frac{{dv}}{{{v^2}}}} &= - \int {\frac{k}{m}} dt\frac{{{v^{ - 2 + 1}}}}{{ - 2 + 1}}\\ &= - \frac{k}{m}t + {C_1} - \frac{1}{v}\\ &= - \frac{k}{m}t + {C_1}\frac{1}{v}\end{aligned}\)

\( = \frac{k}{m}t - {C_1}\) ……. (4)

Substitute the initial value \(v(0) = {v_0}\) in equation (4),

\(\begin{aligned}\frac{1}{{{v_0}}} &= \frac{k}{m}(0) - {C_1}\\{C_1} &= -\frac{1}{{{v_0}}}\end{aligned}\)

Substitute \(C\) value in equation (3)

\(\frac{1}{v} = \frac{k}{m}t + \frac{1}{{{v_n}}}\)

The velocity from the above solution

\(\begin{aligned}v &= \frac{1}{{\left( {\frac{{kt}}{m}} \right) + \left( {\frac{1}{{{v_0}}}} \right)}}\\v &= \frac{1}{{\left( {\frac{{{v_0}k + m}}{{m{v_0}}}} \right)}}\\v(t) &= \left( {\frac{{m{v_0}}}{{{v_0}kt + m}}} \right)\end{aligned}\)

Therefore, the velocity is \(v(t) = \left( {\frac{{m{v_0}}}{{{v_0}kt + m}}} \right)\).

Calculate the position with respect to velocity

The change of position is \(s\) with respect to \(t\) is velocity,

\(\begin{aligned}v(t) &= \frac{{ds}}{{dt}}\\ds &= v(t)dt\end{aligned}\)

Integrate on both sides,

\(\begin{aligned}\int d s &= \int v (t)dt\\\int d s &= \int {\left( {\frac{{m{v_0}}}{{{v_0}kt + m}}} \right)} dt\\s(t) &= m{v_0}\left( {\frac{1}{{{v_0}k}}} \right)\ln \left| {m + {v_0}kt} \right| + {C_2}\end{aligned}\)

\(s(t) = \left( {\frac{m}{k}} \right)\ln \left| {m + {v_0}kt} \right| + {C_2}\) ……. (5)

Substitute the initial value \(s(0) = {s_0}\) in equation (5),

\(\begin{aligned}{s_0} &= \left( {\frac{m}{k}} \right)\ln \left| {m + {v_0}k(0)} \right| + {C_2}\\{s_0} &= \left( {\frac{m}{k}} \right)\ln (m) + {C_2}\\{C_2} = {s_0} - \left( {\frac{m}{k}} \right)\ln (m)\end{aligned}\)

Substitute \({C_2}\) value in equation (5),

\(\begin{aligned}s(t) &= \left( {\frac{m}{k}} \right)\ln \left| {m + {v_0}kt} \right| + {s_0} - \left( {\frac{m}{k}} \right)\ln (m)\\s(t) &= \left( {\frac{m}{k}} \right)\left( {\ln \left( {\left| {m + {v_0}kt} \right| - (m)} \right)} \right) + {s_0}\\s(t) &= \left( {\frac{m}{k}} \right)\ln \left( {\left| {\frac{{m + {v_0}kt}}{m}} \right|} \right) + {s_0}\end{aligned}\)

Therefore, the position is \(s(t) = \left( {\frac{m}{k}} \right)\ln \left( {\left| {\frac{{m + {v_0}kt}}{m}} \right|} \right) + {s_0}\).

Calculate the distance travelled

The distance travelled time 0 to \(t = \) Final position - Initial stage.

The total distance travelled \( = \mathop {\lim }\limits_{t \to \infty } \left( {s(t) - {s_0}} \right)\)

\(\begin{aligned}\mathop {\lim }\limits_{t \to \infty } \left( {s(t) - {s_0}} \right) &= \mathop {\lim }\limits_{t \to \infty } \left( {\left( {\frac{m}{k}} \right)\ln \left( {\left| {\frac{{m + {v_0}kt}}{m}} \right|} \right) + {s_0}} \right) - {s_0}\\ &= \mathop {\lim }\limits_{t \to \infty } \left( {\frac{m}{k}} \right)\ln \left( {\left| {\frac{{m + {v_0}kt}}{m}} \right|} \right)\\ &= \frac{{m\infty }}{k}\\ &= \infty \end{aligned}\)

Therefore, the distance travelled is infinite.