Accelerated motion down a frictionless incline

Steps:

• Acceleration down incline: a = g sin θ = 9.8 m/s² × sin 30° = 9.8 × 0.5 = 4.9 m/s².
• Initial velocity u = 0 m/s (released from rest).
• Distance formula: s = ut + ½ at² = 0 + ½ × 4.9 × (0.8)².
• Calculate: ½ × 4.9 × 0.64 = 2.45 × 0.64 = 1.568 m ≈ 1.6 m.

Why A is correct:

• Matches s = ½ (g sin θ) t² from Newton's second law for constant acceleration on incline.

Why the others are wrong:

• B: Overestimates distance by ignoring sin 30° factor or using incorrect time.
• C: Uses a = g cos 30° (normal force component) instead of parallel acceleration.
• D: Uses full g (9.8 m/s²) as if on vertical drop, not incline.

Final answer: A