Calculating displacement from acceleration (intuitively) If I say acceleration of car is constant at $4\; \rm m/s^2$.
Then isn’t it that it covers $4\; \rm m$ in $1\; \rm s$ with velocity $4\; \rm m/s$.
Then in $2\; \rm s$, the velocity is $8\; \rm m/s$. Therefore, I (erroneously) conclude it covers $12\; \rm m$.
By crosschecking using the $\Delta d=\dfrac12at^2+ut$ equation with $u =0$, I get:
$$\dfrac12(4)(2\times2) = 8\rm \; m$$
Obviously, my first approach is wrong, but I do not know why. Could someone please explain why my initial intuitive approach failed?
 A: If you have constant acceleration, then your velocity vs. time graph will be a linear relationship -- $v(t)=at+v_0$.
The reason your approach doesn't work is because the velocities of $4\;\rm m/s$ and $8\;\rm m/s$ are instantaneous velocities. Even though we write "$\rm m/s$", within that one second, the velocity may actually change. So if an object's instantaneous velocity at some time is $4\;\rm m/s$, that does not mean that it will travel 4 meters one second later, because half a second later, its velocity could be $40\;\rm m/s$, which means it's no longer travelling at that $4\;\rm m/s$ you assumed.
If the velocity remains constant, then it will be true that in after one second, the object travels 4 meters.
However, the presence of acceleration obviously implies no constant velocity.

(yes, in my sketch I accidentally did displacement between $t=1\;\rm s$ and $t=3\;\rm s$ instead of $t=0\;\rm s$ and $t=2\;\rm s$ as you did, but it's the same concept)
A: If your acceleration is constant $a(t) = a$ Then
$v(t) = v_0 + at$ thus $x(t) = \frac{1}{2}at^2 + v_0 + x_0$
If your initial velocity and displacement are zero then $x(2) = 8m$
A: Your intuitive approach was in the right direction. Your mistake was that you used the final velocity after each second to determine the distance, whereas you should have used the average velocity. This is because for each second the velocity is constantly changing e.g. the first second it starts with 0 but gradually increases to 4 m/s
So in the first second the average velocity would be: $\frac{4+0}{2} = 2 m/s$ and the so the distance travelled would be 2m.
The next second the average velocity would be $\frac{4+8}{2} = 6 m/s$ and the distance travelled would be 6m
So the total distance travelled is $2m+6m = 8m$
