Question

Two boats are heading away from shore. Boat 1 heads due north at a speed of 2.98 m/s relative to the shore. Relative to Boat 1, Boat 2 is moving 29.0° north of east at a speed of 1.51 m/s. A passenger on Boat 2 walks due east across the deck at a speed of 1.17 m/s relative to Boat 2. What is the speed of the passenger relative to the shore?

Answer 1

Let us write all the given information in terms of x and y where x is unit vector in East direction and y is unit vector in North direction.

V(Boat 1 relative to shore) = 2.98 m/s due north
= 2.98 y

V(Boat 2 relative to Boat 1) = 29.0° north of east at a speed of 1.51 m/s.
= 1.51 cos 29 x + 1.51 sin 29 y
= 1.32 x + 0.73 y

V (passenger relative to boat 2) = 1.17 m/s due east
= 1.17 x

V (passenger relative to shore) = (passenger relative to boat 2) + V(Boat 2 relative to Boat 1) + V(Boat 1 relative to shore)

= 1.17x + 1.32x +0.73y + 2.98 y
= 2.49 x + 3.71 y
Speed of passenger relative to shore = sqrt (2.49^2 + 3.71^2) = 4.5 m/s

Boat is going at an angle thetha north of east.
Thetha = atan (3.71/2.49) = atan (1.5) = 56 degree

So, Speed of passenger relative to shore is 4.5 m/s at an angle of 56 degree north of east.