Second order differential equation for acceleration Can you help me understand and solve this question:

A bullet is fired vertically upwards with an initial velocity of $u$. Form a second order differential equation for acceleration and by integrating twice find an equation for the displacement $s$ traveled by the bullet in terms of time $t$ from firing.

I cannot wrap my head around this. I haven't done any work, don't know even how to start. All I know is that $v(t)=u-gt$, where
$v(t)$ is the velocity at time $t$, $u$ is the initial velocity and $-g$ is the earth acceleration. From this it follows that when $t = 0$ the velocity $v(t)$ will be equal to the initial velocity $u$; and at a certain point of time, the velocity $v(t)$ will be equal to zero. Do I need to differentiate the $v(t)=u-gt$ equation to get to the equation for the acceleration? And then integrate the acceleration equation twice? I am not even sure that $v(t)=u-gt$ is the right velocity equation. 
Thanks a lot!
 A: Suppose that our vertical axis is directed upwards. The second Newton's law tells us that $$m\ddot y=F.$$ The gravity law (again, thanks to Newton) will say in our case that $F=-mg$, which leads usto the differential equation
$$\ddot y=-g.$$
Suppose also that the initial coordinate of your bullet is $y_0$ and initial velocity is $\dot y(0)=v_0$. The total problem (this is a so-called Cauchy problem)
writes
$$\begin{cases}
\ddot y=-g\\
\dot y(0)=v_0\\
y(0)=y_0
\end{cases}$$
We can integrate the differential equation:
$$\int_0^t \ddot y(s)ds= -\int_0^t g\,ds,$$
which gives us (again, Newton's formula)
$$\dot y(t)-\dot y(0)=-gt,$$
or, in other words,
$$v(t) = \dot y(t) = v_0-gt.$$
Now we can integrate the aabove expression once again:
$$\int_0^t \dot y(s)ds =\int_0^t (v_0-gs)ds,$$which gives
$$y(t)-y(0)=v_0t - gt^2/2,$$
Therefore the displacement $D=v_0t - gt^2/2$ and the general formula of movement
$$y(t) = y_0+v_0t-gt^2/2.$$
If you still have questions, ask in comments.
A: In this simple setup the (differential) equations of motion (depending on time t, in one spatial dimension, like up and down) are
$$ s^\prime(t) = v(t) ,\, s^ {\prime\prime}(t) ) = v^\prime(t) = a(t) $$
In your case $a(t) = -g$ is a constant. The equation you wrote down $v(t)=u-gt$ is just the integral of the last one with initial condition $v(0) = u$. You need to integrate this once more wrt to $t$ and initial condition = 'height of the end of the gun over the floor' to get $s$.
A: Let $t=0$ be the instant when the bullet was fired. Then the velocity as a function of time is, $v(t)=u-gt$, if we assume that a positive value means that it points upwards.
The acceleration is given by, $a(t)=\dfrac{d}{dt}v(t)=-g$.
The displacement is the integral of velocity or the second integral of acceleration. 
Can you take it from here? 
Note that integrating the function $a(t)=-g$ twice will give you two unknown constants in the expression of displacement, but one of them can be determined since you know the initial velocity, $u$.
A: If you assume that the earth's gravitational force is constant along the path of the bullet, the acceleration is constant and equal to $-g = -9.81\frac{m}{s^2}$. Thus, $$
  a(t) = -g \text{.}
$$
Since $a = v' = s''$, you get for the displacement $s$ that $$
  s''(t) = -g, s'(t_0) = v_0
$$
and since we assumed that $g$ doesn't depend on the position, you may wlog add that $s(t_0) = 0$.
Now, this isn't really a differential equation at all, because only one derivative of $s$ appear in it - namely $s''$. You can thus just integrate and get $$
  s'(t) = v(t) = v_0 + \int_{t_0}^t -g \,du = v_0 -g(t-t_0)
$$
and from that (assume $s(t_0) = 0$) $$
  s(t) = \int_{t_0}^t v(u) \,du = v_0(t -t_0) - g\tfrac{1}{2}(t-t_0)^2 \text{.}
$$
