Question

In velocity time graph, area under v-t graph represents :

• A. Displacement
• B. Velocity
• C. Acceleration
• D. Time

Hint:

For a velocity-time (v-t) graph: The {area} under the graph gives the {displacement}. The {slope} of the graph gives the {acceleration}. Remember these two key interpretations, as they are fundamental in kinematics.

Answer 1

The Correct Option is A

Concept: A velocity-time (v-t) graph plots the velocity of an object against time. The physical quantities displacement and acceleration can be determined from this graph. Step 1: Understanding the relationship between velocity, time, and displacement Velocity is defined as the rate of change of displacement, \(v = \frac{\Delta d}{\Delta t}\), where \(\Delta d\) is the change in displacement and \(\Delta t\) is the time interval. For constant velocity, displacement \(d = v \times t\). If velocity is not constant, we consider a small time interval \(dt\) during which the velocity \(v\) can be considered approximately constant. The small displacement \(dd\) during this interval is \(dd = v \cdot dt\). To find the total displacement over a period from time \(t_1\) to \(t_2\), we integrate this expression: \(d = \int_{t_1}^{t_2} v \, dt\). Mathematically, the definite integral \(\int_{t_1}^{t_2} v \, dt\) represents the area under the curve of \(v\) plotted against \(t\), from \(t_1\) to \(t_2\). Step 2: Interpreting the area under the v-t graph Consider a simple case where an object moves with a constant velocity \(v\) for a time \(t\). On the v-t graph, this would be a horizontal line at height \(v\). The area under this line from \(t=0\) to \(t\) would be a rectangle with height \(v\) and width \(t\). Area = height × width = \(v \times t\). Since displacement for constant velocity is also \(v \times t\), the area under the v-t graph is equal to the displacement. This principle extends to any shape under the v-t graph; the area always represents the displacement. Step 3: Distinguishing from slope The slope of a velocity-time graph represents acceleration. Slope is calculated as the change in velocity divided by the change in time (\(\frac{\Delta v}{\Delta t}\)), which is the definition of acceleration. Option (3) Acceleration is incorrect as it is represented by the slope, not the area. Therefore, the area under a velocity-time graph represents displacement.