$(a)$ $A$ car moving with uniform velocity $u$ and uniform acceleration $a$ covers a distance $S$ in time $t$. Draw its velocity-time graph and derive an expression relating all the given physical quantities.
$(b)$ $A$ boy revolves a stone tied to a string $0.7 \, m$ long. Find the distance and displacement covered by the stone in completing two revolutions from the starting point.

(N/A) The velocity-time graph for uniform acceleration is a straight line inclined to the time axis. The area under the velocity-time graph gives the distance covered. By integrating or using the area of a trapezium (or rectangle + triangle),we derive the equation of motion: $S = ut + \frac{1}{2}at^2$.
$(b)$ Given: Radius $r = 0.7 \, m$. Number of revolutions $n = 2$.
Distance is the total path length covered: $\text{Distance} = n \times (2\pi r) = 2 \times 2 \times \frac{22}{7} \times 0.7 = 8.8 \, m$.
Displacement is the shortest distance between the initial and final positions. Since the stone completes two full revolutions,it returns to the starting point. Therefore,$\text{Displacement} = 0 \, m$.