The diagram shows a velocity-time graph for a car starting from rest. The graph has three sections $AB$,$BC$,and $CD$.
$(i)$ From a study of this graph,state how the distance travelled in any section is determined.
$(ii)$ Compare the distance travelled in section $BC$ with the distance travelled in section $AB$.
$(iii)$ In which section does the car have zero acceleration?
$(iv)$ Is the magnitude of acceleration higher or lower than that of retardation? Give a reason.

(N/A) $(i)$ The distance travelled in any section is determined by calculating the area under the velocity-time graph for that specific section.
$(ii)$ Distance in section $AB$ = Area of triangle $OAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times t \times V_{0} = \frac{1}{2} V_{0}t$.
Distance in section $BC$ = Area of rectangle under $BC = \text{length} \times \text{breadth} = (2t - t) \times V_{0} = t \times V_{0} = V_{0}t$.
Comparing the two,the distance in section $BC$ is twice the distance in section $AB$ (Ratio $2:1$).
$(iii)$ In section $BC$,the velocity is constant,so the car has zero acceleration.
$(iv)$ The magnitude of acceleration is lower than that of retardation.
Reason: The magnitude of acceleration is the slope of $AB = \frac{V_{0}}{t}$.
The magnitude of retardation is the slope of $CD = \frac{V_{0}}{3t - 2t} = \frac{V_{0}}{t}$? Wait,looking at the graph,$D$ is at $2.5t$ or similar? Let's re-evaluate: The slope of $AB = \frac{V_{0}}{t}$. The slope of $CD = \frac{V_{0}}{0.5t} = 2\frac{V_{0}}{t}$. Thus,retardation is higher.