Distance from the area of the v-t graph

Problem

The graph in the figure shows the velocity of a car as a function of time. Determine the distance covered in the first $9\;\text{s}$ by calculating the area under the curve.

Solution

Key idea. In a velocity-time graph, the distance travelled is numerically equal to the area between the curve and the time axis. So it suffices to calculate the area of the coloured region.

Reading the graph. The velocity starts at $0$, grows linearly up to $6\;\text{m/s}$ at instant $t = 3\;\text{s}$, stays constant at $6\;\text{m/s}$ until $t = 6\;\text{s}$, then decreases linearly back to $0$ at $t = 9\;\text{s}$. The figure is a trapezium.

Method 1 — decomposition into elementary shapes. A rising triangle, a central rectangle and a falling triangle:

$d = \underbrace{\tfrac{1}{2}\cdot 3\cdot 6}_{9} + \underbrace{3\cdot 6}_{18} + \underbrace{\tfrac{1}{2}\cdot 3\cdot 6}_{9} = 36\;\text{m}$

Method 2 — area of the trapezium. Longer base $B = 9\;\text{s}$, shorter base $b = 3\;\text{s}$ (the constant-velocity stretch), height $h = 6\;\text{m/s}$:

$d = \frac{(B + b)}{2}\, h = \frac{(9 + 3)}{2}\cdot 6 = 36\;\text{m}$

The two methods agree, as they must.

$\ev{d = 36\;\mathrm{m}}$

Note: the average velocity over the whole interval is $\bar v = d/\Delta t = 36/9 = 4\;\text{m/s}$, i.e. the height of the rectangle of equal area and base $9\;\text{s}$.

Links

Topics: Kinematics Concepts: Average speed Skills: Reading graphs