Canoe crossing

From IGCSE mock exams last month, my daughter said one problem in particular bothered her a bit, and that she wasn’t sure she’d gotten it right. Here’s that problem:

So, we sat down to see if we could solve it at home. When reading the problem for the first time, there is a higher tendency to think that the canoe in still water would travel normal to the bank. It’s erroneous, and resolving it further by applying plain geometry or trigonometry right away leads one astray. For a simple looking problem, it hides its complexity in plain sight!

After a bit of brainstorming, she came up with the idea that the trick to getting it right is to first consider the contribution of the current to the distance travelled parallel to the bank. Once we realized this, the rest fell into place. Here’s how:

w = 40 m     # river width
u = 1.8 m/s  # river speed (current)
d = 70 m     # Canoe distance along river from P to Q 
t = 12 s     # Time taken by canoe from P to Q

Canoe is forced to travel the following distance due to current, whose component is included in distance, d (this step is crucial in finding a solution):

dc = u * t => 21.6 m

Canoe distance along river in still water:

d0 = d - dc => 48.4 m

Distance from P to Q (new hypotenuse) in still water:

dn = sqrt(w**2 + d0**2) => 62.79 m

Speed of the canoe in still water:

v = dn / t = 5.23 m/s

Angle to the bank that the canoe was steered in still water:

theta = atan(w / d0) in deg => 39.57 deg

Although the problem does not ask, we can also calculate speed of the canoe under current’s influence:

    vc = d / t => 5.83 m/s

The problem has a real-world flavor to it, and I can think of one in nearshore or coastal areas, where room to maneuver vessels in congested ports is often low, and a bit of back of the envelope calculation helps.