$(a)$ Derive the second equation of motion $S = ut + \frac{1}{2}at^2$ graphically,where the symbols have their usual meanings.
$(b)$ $A$ car accelerates uniformly from $18 \text{ km h}^{-1}$ to $36 \text{ km h}^{-1}$ in $5 \text{ s}$. Calculate the acceleration and the distance covered by the car in that time.

(A) The area under the velocity-time graph represents the distance travelled by the object. As shown in the figure,the area is the sum of the area of the rectangle $OACD$ and the area of the triangle $ABC$.
The area of the rectangle $OACD = \text{length} \times \text{breadth} = OC \times OA = t \times u = ut$.
The area of the triangle $ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AC \times BC = \frac{1}{2} \times t \times (v - u)$.
Since $v = u + at$,we have $(v - u) = at$. Substituting this into the area of the triangle,we get:
Area of triangle $ABC = \frac{1}{2} \times t \times (at) = \frac{1}{2}at^2$.
Therefore,the total distance $S = \text{Area of rectangle } OACD + \text{Area of triangle } ABC = ut + \frac{1}{2}at^2$.
$(b)$ Given:
Initial velocity $u = 18 \text{ km h}^{-1} = 18 \times \frac{5}{18} \text{ m s}^{-1} = 5 \text{ m s}^{-1}$.
Final velocity $v = 36 \text{ km h}^{-1} = 36 \times \frac{5}{18} \text{ m s}^{-1} = 10 \text{ m s}^{-1}$.
Time $t = 5 \text{ s}$.
Acceleration $a = \frac{v - u}{t} = \frac{10 - 5}{5} = \frac{5}{5} = 1 \text{ m s}^{-2}$.
Distance $S = ut + \frac{1}{2}at^2 = (5 \times 5) + \frac{1}{2} \times 1 \times (5)^2 = 25 + 12.5 = 37.5 \text{ m}$.