A sailboat has a mass of 1.50×10^3 kg and is acted on by a force of 2.00×103 N toward the east, while the wind acts behind the sails with a force of 3.00×10^3 N in a direction 45° north of east. Find the magnitude and direction of the resulting acceleration.

Acceleration is 3.09 m/s²

The direction of the acceleration is

N 62.742° E

Explanation:

A sailboat has a mass of 1.50×10^3 kg and a force of 2.00×10^3 N is

acting on it toward the east.

The wind with force of 3.00×10^3N

In a direction 45° north of east act on it.

The resultant force is

C² = A² + B² - 2ABCosc

C² = 2² + 3²- 2(2)(3)cos135

C² = 4 + 9 + 8.485

C² = 13+8.485

C² = 21.485

C = 4.635 * 10^3 N

Let sinb = x

3/x = 4.635/sin135

3/x = 6.555

X= 3/6.555

X = 0.458

But x = sin b

b = sin^-1 0.458

b = 27.258°

90- 27.258 = 62.742°

The direction of the force is

N 62.742° E

Acceleration is F/m

a = 4.635 * 10^3/1.50×10^3

a = 3.09 m/s²

for this one it would be d bump/forearm pass

i'm thinking more like energy