What is the reading of the instrument?

The zero mark of the vernier scale is past $2.3 \text{ cm}$ but has not reached $2.4 \text{ cm}$.
So, the main scale reading is $2.3 \text{ cm}$.

Identify the mark on the vernier scale that perfectly aligns with any mark on the main scale. In the diagram, the $8^{\text{th}}$ mark on the vernier scale aligns. Assuming a least count of $0.01 \text{ cm}$, the vernier scale reading is $8 \times 0.01 \text{ cm} = 0.08 \text{ cm}$.

Total reading = Main scale reading + Vernier scale reading

The correct option is C.

A2: Distance from Speed-Time Graph

Speed is increasing from $t=0 \text{ s}$ to $t=20 \text{ s}$. This forms a triangle on the graph.
The base of the triangle is $20 \text{ s}$ and the height is $10 \text{ m/s}$.
Distance (increasing) = Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height}$

Speed is decreasing from $t=40 \text{ s}$ to $t=50 \text{ s}$. This forms another triangle.
The base of the triangle is $(50 - 40) \text{ s} = 10 \text{ s}$ and the height is $10 \text{ m/s}$.
Distance (decreasing) = Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height}$

Total distance = Distance (increasing) + Distance (decreasing)

The correct option is B.

A3: Distance during Retardation

Initial velocity, $u = 9 \text{ m/s}$
Final velocity, $v = 5 \text{ m/s}$
Time, $t = 10 \text{ s}$

For constant acceleration (or retardation), the distance $s$ can be found using the formula:

The correct option is C.