The area under acceleration-time graph represents:
A. Velocity
B. Distance
C. Displacement
D. Change in velocity
Answer: Option D
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.
A particle performs Simple Harmonic Motion (SHM). If its displacement from the mean position is half of its amplitude, the ratio of its kinetic energy to potential energy is:
Explanation:
In SHM, Total Energy (E) = 1/2 k A^2. Potential Energy (U) at displacement x is 1/2 k x^2. Kinetic Energy (K) is E - U = 1/2 k (A^2 - x^2). Given x = A/2. U = 1/2 k (A/2)^2 = 1/2 k (A^2 / 4) = E/4. Since total energy is conserved, K = E - E/4 = 3E/4. The ratio of Kinetic to Potential Energy (K:U) is (3E/4) / (E/4) = 3:1.
A stone is dropped from a tower. The distance covered by it in the last second of its fall is equal to the distance covered in the first three seconds. The height of the tower is approximately: (g = 10 m/s²)
Explanation:
Distance in first 3 seconds: s₃ = ½gt² = ½×10×9 = 45 m. Let total time be n seconds. Distance in nth second: sₙ = u + g(2n-1)/2 = 0 + 5(2n-1) = 10n - 5. Given sₙ = 45 m ⇒ 10n - 5 = 45 ⇒ n = 5 s. Total height h = ½gn² = ½×10×25 = 125 m. This problem combines equation of motion with logical reasoning about time intervals. Exam tip: For free fall from rest, distance in nth second = 5(2n-1) meters when g=10 m/s². Such numerical problems test application skills in competitive exams.
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