**6.2 Quantity Represented by the Area under a Graph**

**1.**In the speed-time graph,

**(a)**Quantity represented by the gradient of the graph is

**acceleration**or the

**rate of change of speed**.

**(b)**Quantity represented by the area under the graph is

**distance**.

**Example 1:**

Calculate the distance of each of the following graphs.

Distance = Area under the speed-time graph =

**Area of a triangle**
$\begin{array}{l}\text{Distance}=\frac{1}{2}\times base\times height\\ \text{}=\frac{1}{2}\times 7\times 6=21\text{}m\end{array}$

**(b)**Distance = Area under the speed-time graph =

**Area of a rectangle**Distance = Length × Breadth

= 6 × 4 = 24 m

**(c)**Distance = Area under the speed-time graph =

**Area of a trapezium**

$\begin{array}{l}\text{Distance}=\frac{1}{2}\left(a+b\right)h\leftarrow \overline{)\begin{array}{l}\text{Area of trapezium}\\ =\frac{1}{2}\times \text{Sum of the two}\\ \text{parallel sides}\times \text{Height}\end{array}}\\ \text{}=\frac{1}{2}\left(4+6\right)\times 8=40\text{}m\end{array}$

__Combination of Graphs__**Example 2:**

The diagram above shows the speed-time graph of a moving object for 15 seconds.

**(a)**State the length of time, in s, that the particle moves with constant speed.

**(b)**Calculate the rate of change of speed, in ms

^{-2}, in the first 3 seconds.

**(c)**Calculate the average speed of the object in 15 seconds.

Solution:Solution:

**(a)**

Length of time that the particle moves with constant speed

= 9 – 3 =

**6 s**

(b)

(b)

Rate of change of speed in the first 3 seconds

= acceleration = gradient

$\begin{array}{l}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\ =\frac{6-3}{3-0}\\ =1\text{}m{s}^{-2}\end{array}$

**(c)**

Total distance travelled of the object in 15 seconds

= Area under the graph in the 15 seconds

= Area

*P*+ Area*Q*+ Area*R*
$\begin{array}{l}=\left[\frac{1}{2}\left(3+6\right)\times 3\right]+\left[\left(9-3\right)\times 6\right]+\left[\frac{1}{2}\left(15-9\right)\times 6\right]\\ =27+36+18\\ =81\text{}m\end{array}$

Average speed of the object in 15 seconds

$\begin{array}{l}=\frac{\text{Total distance travelled}}{\text{Total time taken}}\\ =\frac{81}{15}=5.4\text{}m{s}^{-1}\end{array}$