Aplane is flying east at 115 m/s. the wind accelerates it at 2.88 m/s^2 directly northwest. after 25.0 seconds what is the direction of the velocity of the plane?

38.5° north of east

Explanation:

Directly northwest is 135° from east.  If we say east is +x and north is +y, then the components of the acceleration are:

ax = 2.88 cos 135° ≈ -2.036 m/s²

ay = 2.88 sin 135° ≈ 2.036 m/s²

And the initial velocity is:

v₀x = 115 m/s

v₀y = 0 m/s

After, 25.0 seconds, the final velocity components are:

vx = at + v₀x

vx = (-2.036)(25.0) + 115

vx ≈ 64.1 m/s

vy = at + v₀y

vy = (2.036)(25.0) + 0

vy ≈ 50.9 m/s

The plane's direction is:

θ = atan(vy / vx)

θ = atan(50.9 / 64.1)

θ ≈ 38.5°

The direction of the plane's velocity is approximately 38.5° north of east.

do you have a formula possibly

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west