#### Answer

1.04 m/s

#### Work Step by Step

Let east be in the x direction and across the river be the y direction.
Let B denote Boat, R the River, and S the Shore. The pair “RS”, for example, represents the river’s motion relative to the shore, which is (-V, 0) where V represents the river’s speed. The river flows west.
$$ \vec{v_{BS}} = \vec{v_{BR}} + \vec{v_{RS}} $$
$$ (v_{BS} cos \theta, v_{BS} sin \theta) m/s = (2.50 cos 45 ^{\circ}, 2.50 sin 45 ^{\circ}) m/s + (-V, 0) $$
We can find the angle $\theta$, the angle of the boat’s velocity relative to the shore, from the book's diagram. The boat is directed at an angle of 22.49$^o$ east of north (that is the inverse tangent of 118 m/285 m), which makes the angle $\theta = 67.51 ^{\circ} $. In other words, the boat moves relative to the shore at an angle of $\theta = 67.51 ^{\circ} $.
Equate the y-components of the velocities to find the speed of the boat relative to the shore.
$$v_{BS} sin \theta = (2.50 m/s) sin 45 ^{\circ} + 0$$
$$ v_{BS} = 1.91 m/s$$
Next, equate the x-components of the velocities to find the speed of the current.
$$v_{BS} cos \theta = (2.50 m/s) cos 45 ^{\circ} - V$$
$$(1.91 m/s) cos 67.51 ^{\circ} = (2.50 m/s) cos 45 ^{\circ} - V$$
$$V = 1.04 m/s$$