The area under the velocity time graph shows

Question

Accepted Answer

The correct option is (2):
When we analyze motion, graphs are powerful tools. A velocity-time graph plots the velocity of an object against time. The fundamental concept here is that the area under a curve in a graph represents the product of the quantities on the two axes. In a velocity-time graph, the y-axis represents velocity ($v$) and the x-axis represents time ($t$). Therefore, the area under this graph will be the product of velocity and time. We know that velocity is defined as the rate of change of displacement, or $v = \frac{ds}{dt}$, where $s$ is displacement. Rearranging this equation, we get $ds = v , dt$. Integrating both sides, we find that $\int ds = \int v , dt$, which means $s = \int v , dt$. The integral of velocity with respect to time represents the total displacement of the object during that time interval. Geometrically, this integral corresponds precisely to the area under the velocity-time curve. Thus, the area under the velocity-time graph gives the displacement of the object.

The incorrect option is (1):
Acceleration is the rate of change of velocity, or $a = \frac{dv}{dt}$. On a velocity-time graph, acceleration is represented by the slope of the graph, not the area under it.

The incorrect option is (3):
Momentum ($p$) is defined as the product of mass ($m$) and velocity ($v$), i.e., $p = mv$. The area under a velocity-time graph does not directly provide momentum; it would require multiplication by mass, which is not represented on the axes of this particular graph.

The incorrect option is (4):
Time is represented on the x-axis of the velocity-time graph. The area under the graph is not time itself, but rather the product of velocity and time, which yields displacement.