The area under a velocity-time graph represents: MCQ with Answer and Explanation

The area under a velocity-time graph represents:
A. Acceleration
B. Displacement
C. Both displacement and distance travelled
D. Distance travelled
Answer: Option C
Solution (By JKSSB Mock Tests)
The area under velocity-time graph gives displacement when direction is considered (signed area), and distance travelled when absolute values are taken (unsigned area). For unidirectional motion, both are equal. Velocity is rate of change of displacement, so integrating velocity over time yields displacement. This graphical interpretation is fundamental in kinematics. Memory tip: Slope of v-t graph = acceleration; area under v-t graph = displacement. Competitive exams often combine graph analysis with conceptual understanding to test depth of knowledge in motion topics.

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Practice More Physics Questions

Question #1
If the length and cross-sectional area of a wire are both doubled, its resistance will:
A. Double
B. Remain unchanged
C. Halve
D. Quadruple

Correct Answer: Option B


Explanation:
Resistance (R) of a wire is given by R = ρL/A, where ρ is resistivity, L is length, and A is cross-sectional area. If L becomes 2L and A becomes 2A, the new resistance is R' = ρ(2L)/(2A). The 2s cancel out, leaving R' = ρL/A = R. Thus, the resistance remains entirely unchanged.

This question belongs to: Science Physics
Question #2
The dimensional formula of Planck's constant is same as that of
A. Angular momentum
B. Force
C. Energy
D. Linear momentum

Correct Answer: Option A


Explanation:
Planck's constant h = E/ν. Dimensions: [ML²T⁻²]/[T⁻¹] = [ML²T⁻¹]. Angular momentum = mvr = [ML²T⁻¹]. Linear momentum [MLT⁻¹], force [MLT⁻²], energy [ML²T⁻²]. So angular momentum matches. This is a frequently asked SSC question.

This question belongs to: Science Physics
Question #3
A car accelerates uniformly from 18 km/h to 36 km/h in 5 seconds. The distance covered is
A. 45 m
B. 37.5 m
C. 75 m
D. 54 m

Correct Answer: Option B


Explanation:
Convert speeds: 18 km/h = 5 m/s, 36 km/h = 10 m/s. Using v = u + at => a = (10-5)/5 = 1 m/s². Distance s = ut + ½at² = 5×5 + ½×1×25 = 25 + 12.5 = 37.5 m. Also using average velocity: (5+10)/2 × 5 = 7.5×5 = 37.5 m. Answer A.

This question belongs to: Science Physics