if a runner is moving 2.88m/s directly north and accelerating at .350 m/s^2 at an angle of -52.0. At the point in the motion where she is running directly east what is the displacement

19.1 m, 52.0°

Explanation:

Given:

v₀ₓ = 0 m/s

v₀ᵧ = 2.88 m/s

vᵧ = 0 m/s

aₓ = 0.350 m/s² cos(-52.0°) ≈ 0.215 m/s²

aᵧ = 0.350 m/s² sin(-52.0°) ≈ -0.276 m/s²

Find: Δx and Δy

First, find time.

vᵧ = aᵧt + v₀ᵧ

0 m/s = (-0.276 m/s²) t + 2.88 m/s

t = 10.4 s

Now find the displacement components Δx and Δy.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (0 m/s) (10.4 s) + ½ (0.215 m/s²) (10.4 s)²

Δx = 11.7 m

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (2.88 m/s) (10.4 s) + ½ (-0.276 m/s²) (10.4 s)²

Δy = 15.0 m

The magnitude of the displacement is:

d² = Δx² + Δy²

d² = (11.7 m)² + (15.0 m)²

d = 19.1 m

The direction of the displacement is:

θ = atan(Δy / Δx)

θ = atan(15.0 m / 11.7 m)

θ = 52.0°

if the forces acting on an object are unbalanced, the object will accelerate in the direction of the net force on the object.

explanation:

this is best explained by having an understanding of newton's second law, which states that f=ma. you see, when the forces are unbalanced, net force (sum of all forces) is not zero. therefore making it necessary for the object to have an acceleration in the direction of the net force.

b.

explanation: