The area under acceleration-time graph represents:
A. Distance
B. Displacement
C. Change in velocity
D. Velocity
Answer: Option C
Solution (By JKSSB Mock Tests)
Acceleration a = dv/dt ⇒ dv = a·dt. Integrating: ∫dv = ∫a·dt ⇒ Δv = area under a-t graph. Slope of v-t gives acceleration; area under a-t gives velocity change. Memory aid: 'a-t graph area = Δv; v-t graph area = displacement'. Graph interpretation skill essential for motion analysis in competitive exams.
Explanation:
Surface tension = force/length. Force dimensions [MLT⁻²], length [L], so [MLT⁻²]/[L] = [MT⁻²]. Also energy per area gives same dimensions: [ML²T⁻²]/[L²] = [MT⁻²].
Explanation:
Convert speeds: 18 km/h = 5 m/s, 36 km/h = 10 m/s. Using v = u + at => a = (10-5)/5 = 1 m/s². Distance s = ut + ½at² = 5×5 + ½×1×25 = 25 + 12.5 = 37.5 m. Also using average velocity: (5+10)/2 × 5 = 7.5×5 = 37.5 m. Answer A.
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