Question

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?

Answer 1

Answer:

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

Answer 2

Answer:

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.

$V^{2} = U^{2} + 2as$

From the question, V = 0$m/s^{2}$, U = ?, a = -8.0$m/s^{2}$ and s = 484m. so that;

      $0^{2} = U^{2} + 2(-8.0)(484)$

         0 = $U^{2}$ - 7744

⇒ $U^{2}$ = 7744

Find the squareroot of both sides,

$U^{2}$ = 88m/s