Relative Motion in Two Dimensions — Question 10

Question 10

A varying current curves the ground track

A river extends from y=0 to y=120 m. Its eastward water velocity depends on position: $u(y)=[1+y/(60\ \mathrm m)]$ m/s. A boat maintains velocity $(0,4)$ m/s relative to the local water at every point. It starts at $(0,0)$. Use ground x east and y north; the water is not one rigid translating frame.

1. Find the crossing time and ground velocity as functions of time.

2. Integrate the eastward velocity to find downstream drift and the ground-track equation x(y).

3. Find the ground acceleration and explain why constant velocity relative to local water does not imply constant ground velocity.

Question 10 - Solution

A varying current curves the ground track

(a) Use the northward component to parameterize the flow.

$$y=4t,\quad\boxed{T=120/4=30\,\mathrm s},\quad\vec v_G=(1+t/15,4)\,\mathrm{m/s}\quad\text{(numerical }t\text{ in s)}$$ $y=(4\ \mathrm{m/s})t$, so crossing takes 30 s. With $\tau=t/(1\ \mathrm s)$, ground velocity is $(1+\tau/15,4)$ m/s. The eastward component increases as the boat enters faster-flowing water.

(b) Integrate the changing eastward component.

$$x=\int_0^t(1+u/15)\,du=t+t^2/30,\quad x(30)=60\,\mathrm m,\quad x(y)=y/4+y^2/(480\,\mathrm m)$$ $x=[\tau+\tau^2/30]$ m, giving x=60 m at arrival. Eliminating time gives $x=y/4+y^2/(480\ \mathrm m)$. The curved track is not a straight line at the initial velocity or at the final velocity. The mean eastward speed over this crossing is 2 m/s.

(c) Differentiate the ground velocity.

$$a_x=\frac{du}{dy}\frac{dy}{dt}=\frac{1\,\mathrm{m/s}}{60\,\mathrm m}(4\,\mathrm{m/s})=\frac1{15}\,\mathrm{m/s^2},\quad a_y=0$$ Ground acceleration is $(1/15,0)$ m/s$^2$. The stipulated boat-relative-water velocity is unchanged, but the local water velocity varies along the boat’s path. Applying a single constant translating-frame velocity to this entire river would ignore that spatial variation.