Two speedboats can each travel at v= 5 m/s with respect to water. one boat crosses the
river of the width l=1000 m by moving strictly perpendicular from point a to point b and
back. water flows at u=0.5 m/s as shown. the second boat is moving along the center of
the river from points c to d and back, distance s=1000 m. a) how the first boat should
move with respect to water in order to move from a to b? b) find times t1 and t2 needed
for each boat to complete the roundtrip.

Consider the travel of the speedboat from A to B and back.
Refer to the first figure.
The boat travels at 5 m/s relative to the water, and the downstream speed of the water is 0.5 m/s.
Therefore,
V₁=5 m/s, u = 0.5 m/s, sinθ = 0.5/5 = 0.1 => θ = arcsin(0.1) = 5.74°.
The boat should travel upstream at 5 m/s, at an angle of 5.74°.

Similarly, return speed from B to A is 5 m/s, at 5.74° upstream.

The horizontal component of velocity from A to B (or vice versa) is 
5cos(5.74°) = 4.975 m/s.

The time required to travel 1000 m is
1000/4.975 = 202.02  = 3.367 min.
The time for the return trip is
t₁ = 2*3.367 = 6.734 min.

Consider travel from C to D and back, as shown in the second figure.
The resultant velocity upstream from C to D is
V₁ = 5 - 0.5 = 4.5 m/s
The time required to travel 1000 m fromC to D is
1000/4.5 = 222.22 s

The resultant velocity downstream from D to C is
V₂ = 5 + 0.5 = 5.5 m/s
The time required to travel fro D to C is 
1000/5.5 = 181.82 s
Total time for the return trip between C and D is
t₂ = 222.22 + 181.82 = 404.04 s = 6.734 min

The first boat should travel upstream at an angle of 5.74°. The time for the return trip between A and B is t₁ = 6.734 min or 404.04 s.

The second boat travels upstream against the current, and downstream with the current. The time for the return trip between C and D is t₂ = 6.734 min or 404.04 s.

Ithink your answer is d