Answer
$-8.3\space km/h$
Work Step by Step
Here we use equation 2.4 to find the average acceleration (a) of the bus.
$a=\frac{V-V_{0}}{\Delta t}-(1)$
First of all, we have to find the final and initial velocities of the bus. The slope of the position-time graph is the velocity of the bus. Each of the three segments of the graph is a straight line, so the bus has a different constant velocity for each part of the trip. Let's take them as $V_{A},V_{B},V_{C}$
We can write,
$V_{A}=\frac{\Delta x_{A}}{\Delta t_{A}}=\frac{24\space km-0\space km}{1\space h-0\space h}=24\space km/h$
$V_{C}=\frac{\Delta x_{C}}{\Delta t_{B}}=\frac{27\space km-33\space km}{3.5\space h-2.2\space h}=-5\space km/h$
In here $V=V_{C},V_{0}=V_{A}$
(1)=>
$a=\frac{-5\space km/h-(24\space km/h)}{3.5\space h}=-8.3\space km/h^{2}$
