The velocity-time graph of a particle moving in a straight line is shown. The displacement of the particle from t=0 to t=4s is: [Graph: triangle from (0,0) to (2,10) to (4,0)]
Explanation:
Displacement = area under velocity-time graph. The graph forms a triangle with base 4 s and height 10 m/s. Area = ½ × base × height = ½ × 4 × 10 = 20 m. Since velocity is always positive, displacement equals distance travelled. Graphical analysis is powerful: area gives displacement, slope gives acceleration. This triangle represents motion with uniform acceleration followed by uniform retardation. Exam tip: For piecewise linear v-t graphs, calculate area of each geometric segment separately. Such graph-based questions assess conceptual clarity in motion analysis.
Explanation:
Newton's law of gravitation: F = G·m₁m₂/r². Thus force is directly proportional to the product of the masses (m₁m₂) and inversely proportional to square of distance (r²). Option D incorrectly states 'directly proportional to square of distance' – it's inverse square. This fundamental law governs celestial mechanics. Memory tip: 'F ∝ m₁m₂ and F ∝ 1/r²'. Competitive exams often test precise wording of physical laws; distractors may reverse proportionality or misstate dependencies. Always recall the exact mathematical form.
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