Q.What physical quantity is represented by the area under a velocity-time graph? Justify your answer using the integral relationship between velocity and position.

[!TLDR] Since v = dx/dt, integrating both sides gives displacement = ∫v dt, which is exactly the area under the v-t graph between the two times. [!ANSWER] The area unde …

The defining relation. Instantaneous velocity is defined as $v = dx/dt$ (Section 2.4), which rearranges to $dx = v\,dt$.

Integrating. To find the net change in position (displacement) between times $t_1$ and $t_2$, we integrate both sides:

Geometric meaning of the integral. The definite integral $\int_{t_1}^{t_2} v\,dt$ is, by the fundamental geometric definition of a definite integral, exactly the signed area enclosed between the curve $v(t)$ and the time axis, between the vertical lines $t = t_1$ and $t = t_2$ — any portion of the curve above the axis (positive $v$) contributes positive area, and any portion below the axis (negative $v$, i.e. motion in the negative direction) contributes negative area.

Conclusion. Combining the two results directly gives

$\Delta x = x_2 - x_1 = \text{area under the } v\text{-}t \text{ graph between } t_1 \text{ and } t_2$ …

Start from v = dx/dt, rearrange and integrate both sides between the two times, and identify the resulting integ …

• Saying the area gives distance travelled rather than displacement (the two coincide only for one-directional motion) …