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°

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