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)] MCQ with Answer and Explanation
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)]
A. 10 m
B. 20 m
C. 30 m
D. 40 m
Answer: Option B
Solution (By JKExamLibrary)
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:
At maximum height, vertical velocity = 0 momentarily, but acceleration = g downward throughout motion. Memory tip: 'Peak: v=0, a=g; gravity never switches off'. Kinematics concept frequently tested to correct misconception that acceleration vanishes at peak.
Explanation:
g' = g (R/(R+h))². If h=R, g' = g (R/2R)² = g/4. Weight = mg' = mg/4, one-fourth. Above surface, g decreases with square of distance from centre.
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