Question

A plane is flying east at 115 m/s. The wind accelerates it at 2.88 m/s^2 directly northwest. After 25.0s, what is the magnitude of the velocity of the plane?

Answer 1

Answer:

81.8 m/s

Explanation:

The initial velocity of the plane is:

$v_0=115 m/s$ (toward east)

So, decomposing along the x- and y- directions:

$v_{x0} = 115 m/s\\v_{y0} = 0$

(we took east as positive x-direction and north as positive y-direction)

The acceleration is

$a=2.88 m/s^2$ (northwest, so the angle with the positive x-direction is 135 degrees)

Decomposing it along the two directions:

$a_x = a cos 135^{\circ} = (2.88 m/s^2)(cos 135^{\circ})=-2.04 m/s^2\\a_y = a sin 135^{\circ} = (2.88 m/s^2)(sin 135^{\circ})=2.04 m/s^2$

So the two components of the velocity after a time t = 25.0 s will be

$v_x = v_{x0} + a_x t = 115 m/s + (-2.04 m/s^2)(25.0 s)=64 m/s\\v_y = v_{y0} + a_y t = 0 m/s + (2.04 m/s^2)(25.0 s)=51 m/s$

So, the magnitude of the velocity of the plane will be

$v=\sqrt[v_x^2+v_y^2}=\sqrt{(64 m/s)^2+(51 m/s)^2}=81.8 m/s$