Why do these two derivations of the formula for impulse contradict? I was trying to derive $I = F\Delta T$ from $p = mv$.
First I converted $v = at$:
$$\begin{align}
p &= mv \\
p &= m \times at \\
p &= ma \times t \\
p &= Ft \\
\Delta p &= F \times \Delta t \\
I &= F\Delta t
\end{align}$$
which seemed to work.
Then I tried $v = \text{displacement}/t$:
$$\begin{align}
p &= mv \\
p &= m \times d/t \\
p &= mdt^{-1} \\
\Delta p &= -mdt^{-2} \times \Delta t \\
I &= -mdt^{-2} \times \Delta t
\end{align}$$
But when I equate the two results:
$$\begin{align}
F\Delta t &= -mdt^-2 \times \Delta t \\
F &= -m \times dt^{-2} \\
ma &= -m \times a
\end{align}$$
I am confused.
 A: The first part of your working is fine, though it could be clarified a bit. Impulse is equal to change of momentum, so if the momentum changes from $p_1$ to $p_2$ then:
$$ I = p_2 - p_1 = mv_2 - mv_1 = m(v_2 - v_1) $$
If the acceleration $a$ is constant for a time $t$ then we can use the SUVAT equation:
$$ v_2 = v_1 + at $$
so:
$$ v_2 - v_1 = at $$
and as you say this gives us:
$$ I = m \times at = ma \times t = Ft $$
And we correctly conclude that impulse is force $\times$ time.
Where you run into problems with the second approach is that if you're applying a constant acceleration the velocity is constantly changing. So although it's true that the average velocity is $d/t$, this is just an average value. For constant acceleration the relationship between velocity and distance travelled is given by another SUVAT equation:
$$ v_2^2 = v_1^2 + 2ad \tag{1} $$
and the distance travelled is given by:
$$ d = v_1 t + \tfrac{1}{2}at^2 \tag{2} $$
If you substitute equation (2) for $d$ into equation (1) you'll get:
$$\begin{align}
 v_2^2 &= v_1^2 + 2a(v_1 t + \tfrac{1}{2}at^2) \\
       &= v_1^2 + 2av_1 t + a^2t^2 \\
       &= (v_1 + at)^2
\end{align}$$
And square rooting both sides gives us back:
$$ v_2 = v_1 + at $$
just as we had in the first approach.
A: $$I =F t \implies I = mv-mu = m(v-u)$$ where $I$ is the impulse due to force.
