Why is this formula for time in respect to acceleration and distance wrong? "A ball is dropped from $d$ $\text{m}$ off the ground and accelerates downward at $a$ $\text{m/s}^{2}$. How much time $t$ does it take for the ball to hit the ground?"
I tried two different methods for finding the formula for $t$. I know the second is correct and the first is incorrect, but I want to know what is wrong with my first method.
Method 1
I created two equations for speed $s$:
$$
s = at \\
s = \frac{d}{t}
$$
Then I set them equal to each other and solved for $t$:
$$
at = \frac{d}{t} \\
t = \frac{d}{at} \\
t^2 = \frac{d}{a} \\
\bbox[2pt,border: 1px solid black]{t = \sqrt{\frac{d}{a}}} \\
$$
Method 2
For this method, I used: $a$ as the acceleration (since it is downwards it will be negative), $s_i$ as the initial speed, $d_i$ as the initial distance from the ground, $A(t)$ as a function of acceleration in respect to time, $S(t)$ as a function of speed in respect to time, and $D(t)$ as a function of distance in respect to time.
$$
s_i = 0 \\
A(t)=-a \\
S(t) = \int A(t)\ dt = -at + s_i = -at \\
D(t) = \int S(t)\ dt = -\frac{a}{2}t^2+d_i \\
D(t) \overset{\mbox{set}}{=} 0 \\
-\frac{a}{2}t^2 = -d_i \\
t^2 = 2\frac{d_i}{a} \\
\bbox[2pt,border: 1px solid black]{t = \sqrt{2\frac{d_i}{a}}}
$$
 A: littleO and the OP are perfectly correct in their reasoning but I feel as though this doesn't quite get to the core of the problem here. For improved notation I will use $v$ for the speed and $x$ for the position. The equation $x = vt$ is derived as follows.
By definition we have that $v = \frac{dx}{dt}$ thus if we take $v$ to be constant for all $t$ then we find
$$x - x_0 = \int_{x(t_0)}^{x(t)} dx = \int_{t_0}^{t} v \; dt = v (t-t_0) \; . \tag{1}$$
We will then arbitrarily choose that an object subject to this motion begins at $x=0$ and $t=0$
thus we arrive at $x=vt.$ It is extremely important to note that the integral $\textbf{(1)}$ only works if we assume constant velocity for all $t$.
The equation $v=at$ is derived similarly by the definition $a=\frac{dv}{dt} \; .$
If we now work under the assumption that $a$ is constant for all $t$ then
$$v - v_0 = \int_{v(0)}^{v(t)} dv = \int_{t_0}^{t} a \; dt = a (t-t_0) \; .\tag{2}$$
Taking $v_0=0$ and $t_0=0$ we come to find that $v=at \; .$ It is extremely important to note that the integral $\textbf{(2)}$ only works if we assume constant acceleration for all $t$.
If we apply $\textbf{(2)}$, we imply that our object accelerates constantly. If this acceleration is non-zero, then the velocity of the object must change. Recall that $\textbf{(1)}$ cannot be used in the case that velocity is non-constant. Equations $\textbf{(1)}$ and $\textbf{(2)}$ are thus inconsistent with one another. In other words they can't describe the same dimension of the same object's motion.
Let me know if I need to explain anything further.
A: Thanks to littleO for pointing this out:
I incorrectly assumed $at = d/t$. $at$ is the speed of the ball at the moment it hits the ground, and $d/t$ is the average speed of the ball.
Since the acceleration is constant, the valid relationship would be $at/2 = d/t$. This correctly implies that $t = \sqrt{2d/a}$
