The kayaker has velocity vector
v = (2.50 m/s) (cos(45º) i + sin(45º) j )
v ≈ (1.77 m/s) (i + j )
and the current has velocity vector
w = (1.25 m/s) (cos(315º) i + sin(315º) j )
w ≈ (0.884 m/s) (i - j )
The kayaker's total velocity is the sum of these:
v + w ≈ (2.65 m/s) i + (0.884 m/s) j
That is, the kayaker has a velocity of about ||v + w|| ≈ 2.80 m/s in a direction θ such that
tan(θ) = (0.884 m/s) / (2.65 m/s) → θ ≈ 18.4º
or about 18.4º north of east.
