Published by:
CGP EDU Academic Team
Published on: September 11, 2026
Match the Column I with Column II
Column I Graph | Column II Haracteristic | ||
(A) | (p) | Has v > 0 and a < 0 throughout | |
(B) | (q) | Has x > 0 throughout and has a point with v = 0 and a point with a = 0 | |
(C) | (r) | Has a point with zero displacement for t >0 | |
(D) | (s) | Has v < 0 and a > 0 |
A
B
C
D
Text Solution
Verified by ExpertsThe correct answer is:
B
To match the graphs from Column I to their respective characteristics in Column II, we analyze the characteristics given.
(A): The graph shows a negative slope from a positive velocity and negative acceleration, indicating a slowing object moving in the positive direction.
(B): The graph represents an object that starts with positive velocity, reaches zero velocity at some point (turnaround point), indicating it has positive displacement throughout with both velocity and acceleration near zero at a point.
(C): The graph shows that the displacement is decreasing but not crossing zero (always positive), suggesting that it matches the last characteristic indicating zero displacement for t > 0.
(D): This graph shows negative displacement, indicating a negative velocity with positive acceleration.
Therefore, the match is Option B.
(A): The graph shows a negative slope from a positive velocity and negative acceleration, indicating a slowing object moving in the positive direction.
(B): The graph represents an object that starts with positive velocity, reaches zero velocity at some point (turnaround point), indicating it has positive displacement throughout with both velocity and acceleration near zero at a point.
(C): The graph shows that the displacement is decreasing but not crossing zero (always positive), suggesting that it matches the last characteristic indicating zero displacement for t > 0.
(D): This graph shows negative displacement, indicating a negative velocity with positive acceleration.
Therefore, the match is Option B.
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