Question

a car traveling on a straight road at 15 m/s accelerates uniformly to a speed of 21 meters per second in 12 seconds he total distance traveled by the car in this 12 second time interval is...

Answer 1

$$
\text{Acceleration:}\\
a=\cfrac{\Delta velocity}{\Delta time}=\cfrac{21-15}{12}=\cfrac{1}{2}\mathbb{\cfrac{m}{s^2}}
$$
Using equation 5 below:
$$
\begin{align}v & = u + at \quad [1] \\s & = ut + \frac{1}{2} at^2 \quad [2] \\s & = \frac{1}{2}(u + v)t \quad [3] \\v^2 & = u^2 + 2as \quad [4] \\s & = vt - \frac{1}{2}at^2 \quad [5] \\\end{align}
$$
Distance traveled is
$$
s= vt - \frac{1}{2}at^2=21\cdot12-\cfrac{1}{2}\cfrac{1}{2}12^2=216\text{ m}
$$
Make sense?

Answer 2

You could also use equation 2 above
$$
s=ut+\cfrac{1}{2}at^2=15\cdot12+\cfrac{1}{2}\cfrac{1}{2}12^2=216\text{ m}
$$
So you could either use final velocity, time and acceleration as I did in the 1st case, or initial velocity, time and acceleration as I did here. Same answer.

Similarly, you could have used equation 3 and bypass the acceleration calculation altogether. This is the most efficient way:
$$
\large
\bf s=\cfrac{1}{2}(u+v)t=\cfrac{1}{2}(15+21)12=216\text{ m}
$$

You don't even need time to calculate distance, using equation 4, except for the calculation of acceleration.

(Note To determine acceleration in the 1st equation way above, I used equation 1.)