A car accelerates from rest at 2 m/s² for 10 s, then moves with constant velocity for 10 s. Total distance covered is
A. 300 m
B. 200 m
C. 400 m
D. 100 m
Answer: Option A
Solution (By JKSSB Mock Tests)
Acceleration phase: s₁ = ½a t² = ½×2×100 = 100 m, v = a t = 20 m/s. Constant velocity phase: s₂ = v×t = 20×10 = 200 m. Total = 300 m. Motion in two parts.
Explanation:
Time for to-and-fro: t = 2d/v ≥ 0.1 s ⇒ d ≥ vt/2 = 340×0.1/2 = 17 m. Minimum distance for brain to distinguish echo from original sound. Memory tip: 'd_min = v×persistence/2'. Application of sound reflection with human perception limit, common in competitive exam numericals.
No comments yet. Be the first to start the discussion!