Area Between Velocity Curves

I just came across a calculus word problem about integration; specifically, the area between two curves.

Two runners, starting at the same location, run along a straight road for 1 hour. The velocity of one runner is \(\boldsymbol{v_1\left(t\right) = 7t}\) and the velocity of the other runner is \(\boldsymbol{v_2\left(t\right) = 10\sqrt{t}}\). Assume \(\boldsymbol{t}\) is measured in hours and the velocities \(\boldsymbol{v_1\left(t\right) \text{ and } v_2\left(t\right)}\) are measured in km/hr. Determine the area between the curves \(\boldsymbol{y = v_1\left(t\right) \text{ and } y = v_2\left(t\right) \text{, for } 0 \le t \le 1}\). Interpret the physical meaning of this area.

We start by drawing a graph:

We might be tempted to think that we can get the solution to the problem just by looking at the graph. After all, the runners start at the same time and from the same place, and after an hour of running they are, according to the graph \(\boldsymbol{3 km}\) apart, right?

\begin{align*}
\text{One runner:} \quad y &= 7\left(1\right) = 7 \\
\text{The other runner:}\quad y &= 10\sqrt{1} = 10
\end{align*}

Wrong! The graph’s x-axis shows time, but the y-axis is not position but cumulative velocity. While the velocity of one runner is linear, at a constant rate of \(\boldsymbol{7 km/hr}\), the other runner starts faster but his/her velocity is decreasing with time.

Remember that velocity is the derivative of position. When we integrate a velocity function we obtain a position function. Integrating the area between the curves (shaded area in the above graph), the result is the distance between the runners, equivalent to the position of one runner minus the position of the other runner.

\[\int_0^1 \left(10\:t^\frac{1}{2}\: -\: 7\:t\right)\, dt = \left[\frac{20}{3}t^\frac{3}{2}\: -\: \frac{7}{2}t^2\right]_{0}^{1} = \left(\frac{20}{3}\:-\:\frac{7}{2}\right)\:-\:0 = \frac{19}{6} \approx 3.17\:km\]

The physical meaning of the area between the curves is the distance between the runners.