Let's say the car's rate = c, the car's time = d, the boat's rate = b, and the boat's time = a.

From that, we have

Car's Distance = cd
Boat's Distance = ab

and we want |cd - ab|.

S1

30*d = 30*(a - 3)

30d = 30a - 90

d = a - 3

c = 30

b = 30

so

|cd - ab| => |30*(a - 3) - 30*a| => |30a - 90 - 30a| => 90; SUFFICIENT.

You can also see this conceptually by imagining each vehicle. When the boat completes the journey, it and the car have traveled the SAME mileage, since they've each gone for x hours @ 30 mph. At that point, the car has to make up the remaining distance. It takes 3 hours @ 30 mph to do so, so the car travels an extra 90 miles, making the car's distance 90 miles longer.

S2

60*2 = 15*2

This tells us the distance for each: the car has to travel 120 miles (60 mph * 2 hours) and the boat has to travel 30 miles (15 mph * 2 hours). So we have each distance, and can find the different: 120 - 30; SUFFICIENT.