Key physics concepts: Velocity vectors, Distance/speed/time relations Key mathematical concepts: Trigonometry, Pythagorean triples, Vector addition/subtraction.Recall that velocity is a **vector** quantity, and so must have **magnitude** and **direction**, both of which we need to find. The resultant velocity (the speed and direction of the boat as seen by an observer on the bank) is the sum of two components: the water current (known - 5 m/s downstream) and the velocity of the boat relative to the water (unknown - the quantity we want to find!) To cross the river by the shortest distance, we want our resultant velocity vector to be perpendicular to the two banks. Using the fact that v = d/t, we can calculate the magnitude of the vector to be: v = d/t = 72/6 = 12 m/s. Now we have the resultant vector and the current vector, so we can **subtract** the current vector from the resultant to find the velocity of the boat relative to the water. Here it is crucial to draw a diagram (see whiteboard). To subtract two vectors, we put their tails together and draw an arrow from the tip of the vector being subtracted to the tip of the vector being subtracted from (see whiteboard). Given the triangle is right-angled, we can use pythagorus' theorem: a^{2 }+ b^{2} = c^{2 }to calculate the length of this vector (which is the hypotenuse). This gives c = 13 m/s (note the integer result as the values form a pythagorean triple). To find the direction, we can use any of the SOHCAHTOA relations, eg: angle = arcsin(5/13) = 22.6 degrees. So the final answer is: 13 m/s, 22.6 degrees upstream. (Possible extensions: finding a different side of the triangle/ how would you cross the river in the shortest time with a given veloceity and where would you end up?)