The accelerating force of the wind on a small 200-kg sailboat is 707 N northeast. If the drag of the keel is 500 N acting west, what is the acceleration of the boat? a. 1.5 m/s^2 35 degrees north of east
b. 2.5 m/s^2 55 degrees north of west
c. 3.0 m/s^2 35 degrees north of east
d. 4.3 m/s^2 55 degrees north of west
e. 1.5 m/s^2 55 degrees north of west

B

Explanation:

Taking the accelerating force of the wind into cognisance. We can represent it as the sum of two components:

707 /√2 = 500 N acting north, and

707 /√ 2 = 500 N acting east.

The reason is because it is said to be acting NorthEast wards, so it's acting both to the North and to the East. The eastward component will then cancel that of the westward component of the drag of the keel.

Essentially, the sum of two forces that is acting on the boat will then be equal to 500 N north.

To get the acceleration of the boat, we then divide the force by the given mass such that

500/200 = 2.5 m/s^2 northward.

kepler used the data from tycho blahe (his mentor) to construct the laws of planetary motion.

if they have a finite resultant