-2

I understand why we have these two terms in this equation $ v_0t $ and $ (1/2)at^2 $. The thing I don't understand is the area under velocity vs. time graph of the first term.

I get that $vt = d$ but why do we say "area"? For example when our initial velocity is 10 m/s and we want to know how much distance gets travelled in 4 seconds we say

10 m/s * 4 s = 40 m.

But why is it said that this is an area? The units are even cancelling out so why is this considered area?

Qmechanic
  • 220,844
Shadow sparkle
  • 43

1 Answers1

3

Any integral under a curve $f(x)$ can be considered an area, with units of $f(x)\cdot x$. It means it is visually an area in the generic sense with respect to the graph, i.e.

The integral is an accumulation of the function $f(x)$ over the interval $\Delta x = (b-a)$. It does not refer to physical area, e.g. $\mathrm{m}^2$ or $\mathrm{in}^2$.

RC_23
  • 12,766
  • 2
  • 17
  • 53