Question

The area under the velocity-time graph of a particle is equal to its

• A. acceleration
• B. displacement
• C. speed
• D. velocity
• E. angular velocity

Hint:

Use the "D-V-A" rule for graphs:

• \textbf{Slope} of Displacement $\to$ Velocity; Slope of Velocity $\to$ Acceleration.
• \textbf{Area} under Acceleration $\to$ Velocity Change; Area under Velocity $\to$ Displacement.

Answer 1

The Correct Option is B

Concept:
Physics - Kinematics and Graphical Analysis.
Step 1: Define velocity.
Velocity ($v$) is the rate of change of displacement ($s$) with respect to time ($t$): $$ v = \frac{ds}{dt} $$
Step 2: Relate the terms to integration.
From the definition, we can write $ds = v \ dt$.
To find the total change in displacement, we integrate: $$ \Delta s = \int v \, dt $$
Step 3: Interpret integration graphically.
In calculus, the definite integral $\int_{t_1}^{t_2} v(t) \ dt$ represents the geometric area under the curve of the function $v(t)$ between the vertical lines $t = t_1$ and $t = t_2$.
Step 4: Conclusion.
Therefore, the total area enclosed between the velocity-time curve and the time axis directly represents the displacement of the particle.