Draw a velocity-time graph for a body which starts to move with initial velocity $u$ under a constant acceleration $a$ for time $t$. Using this graph,derive an expression for the distance covered $S$ in time $t$.

(N/A) The velocity-time graph for a body moving with constant acceleration $a$ and initial velocity $u$ is shown in the figure.
The distance covered $S$ by the object in time $t$ is equal to the area under the velocity-time graph.
The area under the graph is the sum of the area of the rectangle $OACD$ and the area of the triangle $ABC$ on top of it.
The area of the rectangle $OACD$ is given by: $\text{Area} = \text{length} \times \text{breadth} = t \times u = ut$.
The area of the triangle $ABC$ is given by: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times t \times (v - u)$.
Since the total distance $S$ is the sum of these areas,we have:
$S = ut + \frac{1}{2} \times t \times (v - u)$ $....(1)$
From the first equation of motion,we know that $v = u + at$,which implies $v - u = at$.
Substituting $(v - u) = at$ into equation $(1)$,we get:
$S = ut + \frac{1}{2} \times t \times (at)$
$S = ut + \frac{1}{2}at^{2}$