A398-kg boat is sailing 14.0° north of east at a speed of 1.50 m/s. 23.0 s later, it is sailing 33.0° north of east at a speed of 4.40 m/s. during this time, three forces act on the boat: a 32.8-n force directed 14.0° north of east (due to an auxiliary engine), a 20.5-n force directed 14.0° south of west (resistance due to the water), and (due to the wind). find the (a) the magnitude and (b) direction of the force . express the direction as an angle with respect to due east.

Explanation:

We shall represent speed in vector form

First speed

v₁ = 1.5 cos 14 i + 1.5 sin 14 j

= 1.455 i + 0.363 j

v₂ = 4.4 cos 33 i + 4.4 sin 33 j

= 3.69 i + 2.39 j

v₂ - v₁

3.69 i + 2.39 j - 1.455 i - 0.363 j

= 2.235 i + 2.027 j

acceleration

=  v₂ - v₁ / time

= ( 2.235 i + 2.027 j  ) / 23

= .097 i + .088 j

force = mass x acceleration

= 398 x ( .097 i + .088 j )

= 38.6 i + 35.02 j

Magnitude of force F

F² = 38.6² + 35.02²

F = 52.11 N

Tan θ = 35.02 / 38.6

θ = 42° north of east.

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..