Question

Area under velocity- time graph of a particle is equal to

Hint:

Graph relationships:
- Slope of $x-t$ graph = velocity
- Slope of $v-t$ graph = acceleration
- Area under $a-t$ graph = change in velocity
- Area under $v-t$ graph = displacement

Answer 1

Step 1: Understanding the Concept:
Kinematic graphs relate position, velocity, and acceleration over time. The mathematical relationship between velocity and position is that velocity is the derivative of position with respect to time ($v = dx/dt$). Conversely, position change is the integral of velocity over time.

Step 2: Key Formula or Approach:
The definition of velocity is $v = \frac{dx}{dt}$.
Rearranging this gives $dx = v \cdot dt$.
To find the total change in position ($\Delta x$) between time $t_1$ and $t_2$, we integrate:
\[ \Delta x = \int_{t_1}^{t_2} v(t) \ dt \]

Step 3: Detailed Explanation:
In calculus, the definite integral of a function over an interval geometrically represents the "area under the curve" of that function plotted on a graph.
Therefore, computing the integral $\int v \, dt$ is mathematically identical to finding the area bounded by the velocity-time ($v-t$) curve and the time axis.
The physical quantity represented by $\Delta x$ (change in position) is defined as displacement.
(Note: If the area is calculated considering parts below the time axis as negative, it yields displacement. If the absolute area is summed, it yields total distance traveled. Typically, "area under the graph" implies the signed integral, thus displacement).

Step 4: Final Answer:
The area under a velocity-time graph equals the displacement.