Graph-Based Kinematics MCQs for Class 11
Graph-Based Kinematics MCQs for Class 11 Physics
Class: 11
Subject: Physics
Section: Kinematics
Topic: Graph-Based Kinematics MCQs
Board: CBSE Board Examination (NCERT Based)
Subject: Physics
Section: Kinematics
Topic: Graph-Based Kinematics MCQs
Board: CBSE Board Examination (NCERT Based)
Instructions: These Multiple Choice Questions (MCQs) are designed strictly as per the NCERT syllabus, making them ideal for CBSE Class 11 board examination standards.
Q1. The slope of a displacement–time graph represents:
Answer: B) Velocity
Slope = ΔDisplacement / ΔTime, which by definition gives velocity. A steeper slope indicates higher velocity.
Slope = ΔDisplacement / ΔTime, which by definition gives velocity. A steeper slope indicates higher velocity.
Q2. Area under a velocity–time graph gives:
Answer: B) Displacement
The integral of velocity over time equals displacement. Positive and negative areas indicate direction.
The integral of velocity over time equals displacement. Positive and negative areas indicate direction.
Q3. A horizontal line on a velocity–time graph indicates:
Answer: B) Constant velocity
Slope of v–t graph = acceleration. Horizontal line → slope zero → acceleration zero → constant velocity.
Slope of v–t graph = acceleration. Horizontal line → slope zero → acceleration zero → constant velocity.
Q4. Slope of a velocity–time graph represents:
Answer: C) Acceleration
Acceleration = rate of change of velocity, which is graphically the slope of v–t graph.
Acceleration = rate of change of velocity, which is graphically the slope of v–t graph.
Q5. If slope of v–t graph is negative, motion is:
Answer: B) Retarded
Negative slope → negative acceleration → velocity decreasing → retardation.
Negative slope → negative acceleration → velocity decreasing → retardation.
Q6. A straight line through origin in displacement–time graph shows:
Answer: B) Uniform velocity
Straight line → constant slope → constant velocity.
Straight line → constant slope → constant velocity.
Q7. Curved displacement–time graph indicates:
Answer: C) Accelerated motion
Changing slope means velocity changing → acceleration present.
Changing slope means velocity changing → acceleration present.
Q8. Zero area under v–t graph implies:
Answer: B) Zero displacement
Net area = displacement. If positive and negative cancel → net displacement zero.
Net area = displacement. If positive and negative cancel → net displacement zero.
Q9. Steeper slope in s–t graph means:
Answer: B) Higher velocity
Greater slope → larger displacement change per time → higher velocity.
Greater slope → larger displacement change per time → higher velocity.
Q10. Area under acceleration–time graph gives:
Answer: A) Velocity change
∫a dt = change in velocity.
∫a dt = change in velocity.
Q11. Constant acceleration on v–t graph is shown by:
Answer: B
Straight slanted line → constant slope → constant acceleration.
Straight slanted line → constant slope → constant acceleration.
Q12. Object at rest on s–t graph:
Answer: B
No change in displacement → rest.
No change in displacement → rest.
Q13. Increasing slope in s–t graph means:
Answer: B
Slope rising → velocity increasing → acceleration.
Slope rising → velocity increasing → acceleration.
Q14. Uniform deceleration on v–t graph:
Answer: B
Negative constant slope → uniform retardation.
Negative constant slope → uniform retardation.
Q15. Instantaneous velocity equals slope of:
Answer: B
Tangent slope on s–t graph → instantaneous velocity.
Tangent slope on s–t graph → instantaneous velocity.
Q16. Distance from v–t graph:
Answer: B
Total area ignoring sign → distance.
Total area ignoring sign → distance.
Q17. Parabolic s–t graph indicates:
Answer: B
s ∝ t² → uniformly accelerated motion.
s ∝ t² → uniformly accelerated motion.
Q18. Velocity decreasing uniformly → v–t area shape:
Answer: B
Linear drop forms triangular area.
Linear drop forms triangular area.
Q19. Zero slope in a–t graph means:
Answer: A
Horizontal line → constant acceleration.
Horizontal line → constant acceleration.
Q20. Negative displacement area lies:
Answer: B
Below time axis → negative displacement.
Below time axis → negative displacement.
Q21. Turning point on s–t graph → velocity:
Answer: B
Slope zero at turning point.
Slope zero at turning point.
Q22. Uniform motion → s–t graph:
Answer: B
Q23. Acceleration sign from v–t slope:
Answer: C
Q24. Displacement maximum when area is:
Answer: A
Q25. Speed constant → v–t graph:
Answer: A
Q26. Increasing negative slope →
Answer: A
Q27. Zero area under a–t graph:
Answer: A
Q28. Velocity intercept on v–t graph gives:
Answer: A
Q29. Area rectangle in v–t graph:
Answer: A
Q30. Most useful graph for displacement:
Answer: C
Direct reading of displacement.
Direct reading of displacement.