6. a pilot stops a plane in 484 m using a constant acceleration of -8.0 m/s2. how fast was the plane
moving before braking began?

88 m/s

Explanation:

Given:

v = 0 m/s

Δx = 484 m

a = -8.0 m/s²

Find: v₀

v² = v₀² + 2a(x − x₀)

(0 m/s)² = v₀² + 2 (-8.0 m/s²) (484 m)

v₀ = 88 m/s

The initial velocity of the plane was 88 m/s.

Explanation:

The velocity of the plane before braking began is termed its "initial velocity". To calculate this, we apply Newton's third law of motion to the given question.

From the question, V = 0, U = ?, a = -8.0 and s = 484m. so that;

         0 = - 7744

⇒ = 7744

Find the squareroot of both sides,

= 88m/s

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