Relative Motion in Two Dimensions — Question 1

Question 1

Pointing straight across does not eliminate downstream drift

A river is 120 m wide from south bank to north bank and flows east at 3 m/s. A boat heads due north at 4 m/s relative to the water. 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 boat’s ground velocity, speed and direction east of north.

2. Find the crossing time and downstream drift.

3. Why is dividing the river width by the total ground speed incorrect for the crossing time?

Question 1 - Solution

Pointing straight across does not eliminate downstream drift

(a) Add perpendicular velocity components.

$$\vec v_{B/G}=(0,4)+(3,0)=(3,4)\,\mathrm{m/s},\quad\boxed{v=5\,\mathrm{m/s},\quad\theta=\tan^{-1}(3/4)=36.870^\circ}$$ $\vec v_{B/W}=(0,4)$ and $\vec v_{W/G}=(3,0)$ m/s, so $\vec v_{B/G}=(3,4)$ m/s. Ground speed is 5 m/s and its direction is $\tan^{-1}(3/4)\approx36.870^\circ$ east of north.

(b) Use the component across the banks.

$$T=W/v_y=120/4=30\,\mathrm s,\quad\Delta x=v_xT=3(30)=90\,\mathrm m$$ Time is $120/4=30$ s. Eastward drift is $3(30)=90$ m. The ground displacement is $(90,120)$ m, length 150 m, matching the ground speed times time.

(c) Match distance with the corresponding rate.

$$s=\sqrt{90^2+120^2}=150\,\mathrm m,\quad T=s/v=150/5=30\,\mathrm s\ne120/5$$ The 120 m width is a northward component, not the length of the slanted ground path. Dividing it by 5 m/s would give 24 s and only 96 m of northward travel. Use width divided by northward velocity, or full path length divided by full speed.