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Physics · MOTION IN A STRAIGHT LINE
हिन्दीMatch List-I with List-II regarding graph features in one-dimensional motion: List-I List-II A Position-time curve concave upwards (\( d^2x/dt^2 > 0 \)) I Net displacement over time interval B Position-time curve concave downwards (\( d^2x/dt^2 < 0 \)) II Motion with negative acceleration turning back after stopping C v-t graph with negative slope crossing time axis III Motion with positive acceleration (\( a > 0 \)) D Area under v-t graph between \( t_1 \) and \( t_2 \) IV Motion with negative acceleration (\( a < 0 \)) Choose the correct answer from the options given below:
Match List-I with List-II regarding graph features in one-dimensional motion:
| List-I | List-II | ||
|---|---|---|---|
| A | Position-time curve concave upwards (\( d^2x/dt^2 > 0 \)) | I | Net displacement over time interval |
| B | Position-time curve concave downwards (\( d^2x/dt^2 < 0 \)) | II | Motion with negative acceleration turning back after stopping |
| C | v-t graph with negative slope crossing time axis | III | Motion with positive acceleration (\( a > 0 \)) |
| D | Area under v-t graph between \( t_1 \) and \( t_2 \) | IV | Motion with negative acceleration (\( a < 0 \)) |
Choose the correct answer from the options given below:
View Answer & Solution
Correct answer · Option A
A-(III), B-(IV), C-(II), D-(I)
Explanation
Concave upward x-t curve indicates positive acceleration (A-III). Concave downward x-t curve indicates negative acceleration (B-IV). Negative slope v-t graph crossing zero represents motion turning back under negative acceleration (C-II). Area under v-t graph gives displacement (D-I).
