Relative Motion in Two Dimensions — Question 2

Question 2

Cancel current to land directly opposite

A river is 120 m wide and flows east at 3 m/s. A boat can maintain speed 5 m/s relative to the water and must land directly north of its starting point. Use fixed east-positive x and north-positive y axes. Vector velocities satisfy $\vec v_{A/G}=\vec v_{A/B}+\vec v_{B/G}$ when expressed in the same axes.

1. Find the necessary water-relative velocity and heading west of north.

2. Find the crossing time.

3. Compare with the minimum-time crossing when downstream drift is allowed. Explain why the two optimal headings differ.

Question 2 - Solution

Cancel current to land directly opposite

(a) Cancel the eastward ground component.

$$v_{B/W,x}=-3,\quad v_{B/W,y}=\sqrt{5^2-3^2}=4\quad\mathrm{m/s},\quad\boxed{\theta=36.870^\circ\text{ west of north}}$$ The boat needs $v_{B/W,x}=-3$ m/s. Its fixed-speed condition gives $v_{B/W,y}=+\sqrt{25-9}=4$ m/s, choosing the northward root. Heading is $\tan^{-1}(3/4)\approx36.870^\circ$ west of north.

(b) Use the remaining northward speed.

$$\vec v_{B/G}=(-3,4)+(3,0)=(0,4)\,\mathrm{m/s},\quad T=120/4=30\,\mathrm s$$ The ground velocity is $(0,4)$ m/s and crossing takes $120/4=30$ s. Some of the available water-relative speed is devoted to opposing the current.

(c) Maximize the northward component for minimum time.

$$v_{y,\max}=5\,\mathrm{m/s}\ \Rightarrow\ T_{\min}=120/5=24\,\mathrm s,\quad\Delta x=3(24)=72\,\mathrm m$$ Heading due north gives northward speed 5 m/s and time 24 s, but drift is $3(24)=72$ m east. Directly opposite landing constrains the ground direction; minimum time alone does not. The faster crossing and zero-drift crossing solve different optimization problems.