Amit & Deepali Sinha Foundation

The Boat That Never Stops

Irodov 1.100DifficultyUniqueness

Problem Irodov 1.100

A motorboat of mass mm crosses a lake at speed v0v_0 when, at the instant t=0t=0, its engine is shut off. From that moment the water resists the boat’s motion with a force proportional to its velocity, F=−rv\vect F = -r\vect v, for some constant rr. Find:

  1. how long the motorboat continues to move with the engine off;

  2. the boat’s velocity as a function of the distance it has covered since the engine was shut off, together with the total distance covered before it comes to a complete stop;

  3. the motorboat’s mean velocity over the time interval, starting at t=0t=0, during which its velocity falls to 1/η1/\eta of its initial value.

Given
  • mm
  • v0v_0
  • F=−rv\vect{F} = -r\vect{v}
Find
Find the time to stop; vv as a function of the distance ss covered and the total distance stotals_{\text{total}}; and the mean velocity ⟨v⟩\langle v\rangle while vv falls by a factor η\eta

Solution

Step 1Newton’s second law along the line of motion

The boat moves along a straight line, so drop the vector notation: let v(t)v(t) be the scalar velocity in the direction of motion and ss the distance covered since t=0t=0. The resistive force always opposes the velocity, so along the line of motion Newton’s second law reads

mdvdt=−rv.m\frac{dv}{dt} = -rv.
Step 2How the speed decays with time

Separate the variables and integrate from t=0t=0, where v=v0v=v_0, to a general time tt:

∫v0vdv′v′=−rm∫0tdt′.\int_{v_0}^{v}\frac{dv'}{v'} = -\frac{r}{m}\int_0^{t} dt'.

Both integrals are elementary, and evaluating each at its own limits leaves

ln⁡vv0=−rmt.\ln\frac{v}{v_0} = -\frac{r}{m}t.

Exponentiating both sides gives the speed as a function of time:

v=v0e−rt/mv = v_0 e^{-rt/m}
(1)
Speed against time, illustrative values v_0=4, r/m=1, η=2 (arbitrary units, not the problem's own symbols). The curve approaches the t-axis asymptotically (dotted) without touching it; the marked point is v=v_0/η at t=t^*, used in part (c).

Speed against time, illustrative values v0=4v_0=4, r/m=1r/m=1, η=2\eta=2 (arbitrary units, not the problem’s own symbols). The curve approaches the tt-axis asymptotically (dotted) without touching it; the marked point is v=v0/ηv=v_0/\eta at t=t∗t=t^{*}, used in part (c).

Step 3Trading time for distance

To get vv as a function of the distance ss covered, rewrite the acceleration with the chain rule, since v=ds/dtv = ds/dt:

dvdt=dvdsdsdt=vdvds.\frac{dv}{dt} = \frac{dv}{ds}\frac{ds}{dt} = v\frac{dv}{ds}.

Substitute into the equation of motion:

mvdvds=−rv.mv\frac{dv}{ds} = -rv.

so

m dv=−r ds.m\,dv = -r\,ds.

Integrate from s=0s=0, v=v0v=v_0, to a general ss:

m∫v0vdv′=−r∫0sds′.m\int_{v_0}^{v} dv' = -r\int_0^{s} ds'.

Both sides are integrals of a constant, so each evaluates to its integrand times the length of its own interval:

m(v−v0)=−rs.m(v-v_0) = -rs.

Dividing through by mm and solving for vv gives the speed as a function of the distance covered.

v=v0−rmsv = v_0 - \frac{r}{m}s
(2)
Speed against distance covered, same illustrative values as the v(t) plot above (v_0=4, r/m=1). Unlike the time graph, this relation is exactly linear and reaches zero at a finite distance, marked s_total.

Speed against distance covered, same illustrative values as the v(t)v(t) plot above (v0=4v_0=4, r/m=1r/m=1). Unlike the time graph, this relation is exactly linear and reaches zero at a finite distance, marked stotals_{\text{total}}.

Step 4The total distance covered

Unlike v(t)v(t), which only approaches zero, v(s)v(s) genuinely reaches zero at a finite ss. Setting v=0v=0 in (2) names that distance:

0=v0−rmstotal.0 = v_0 - \frac{r}{m}s_{\text{total}}.

Solving for the stopping distance gives the second half of part (b).

stotal=mv0rs_{\text{total}} = \dfrac{m v_0}{r}
(3)
Check

Dimensions: [m][v0]/[r]=kg⋅(m/s)/(kg/s)=m[m][v_0]/[r] = \mathrm{kg}\cdot(\mathrm{m/s})/(\mathrm{kg/s}) = \mathrm{m}, a length, as required. Also, a larger resistance rr should shorten the stopping distance and a larger mass mm should lengthen it (more inertia carries the boat further); both appear exactly where that requires, rr in the denominator and mm in the numerator.

Step 5Locating the point where the speed has fallen η\eta-fold

Part (c) asks about the interval from t=0t=0 until the speed has fallen to v0/ηv_0/\eta. Impose that speed on the time law (1), calling the instant t∗t^{*}:

v0η=v0e−rt∗/m.\frac{v_0}{\eta} = v_0 e^{-rt^{*}/m}.

Cancel v0v_0 from both sides and take the logarithm of what is left:

rt∗m=ln⁡η.\frac{rt^{*}}{m} = \ln\eta.

Imposing the same speed on the distance law (2) names the distance x∗x^{*} covered by that instant:

v0η=v0−rmx∗.\frac{v_0}{\eta} = v_0 - \frac{r}{m}x^{*}.

Solving each of the last two displays for the quantity it names, and collecting the two over a common denominator in the second,

t∗=mrln⁡η,x∗=mv0r⋅η−1ηt^{*}=\frac{m}{r}\ln\eta, \qquad x^{*}=\frac{mv_0}{r}\cdot\frac{\eta-1}{\eta}
(4)
Step 6The mean velocity over that interval

The mean velocity over [0,t∗][0,t^{*}] is the distance covered divided by the elapsed time, both of them supplied by (4):

⟨v⟩=x∗t∗=mv0r⋅η−1ηmrln⁡η.\langle v\rangle = \frac{x^{*}}{t^{*}} = \frac{\dfrac{mv_0}{r}\cdot\dfrac{\eta-1}{\eta}}{\dfrac{m}{r}\ln\eta}.

The factor m/rm/r cancels, leaving

⟨v⟩=v0 η−1ηln⁡η\langle v\rangle = v_0\,\frac{\eta-1}{\eta\ln\eta}
(5)
Remark

As η→1+\eta\to1^{+} (the speed barely changes over the interval), write η=1+ε\eta=1+\varepsilon with ε→0+\varepsilon\to0^{+}: then (η−1)/(ηln⁡η)→1−ε/2+O(ε2)→1(\eta-1)/(\eta\ln\eta)\to1-\varepsilon/2+O(\varepsilon^2)\to1, so ⟨v⟩→v0\langle v\rangle\to v_0, exactly as it should for an interval over which the speed is nearly constant.

  • (a)

    The speed only approaches zero asymptotically, (1) and the v(t)v(t) plot; formally t→∞t\to\infty.

  • (b)

    The speed falls linearly with distance, (2) and the v(s)v(s) plot, reaching zero at the finite stopping distance (3).