Newton formulated his Second Law as such:
$$\sum{\vec{F}} = \frac{\delta \vec{p}}{\delta t}$$
and of course, $\vec{p} = m \vec{v}$.
Why is it that if the net force $\sum \vec{F}$ is constant (which implies that the rate of change of momentum is constant), then
$$\frac{\delta \vec{p}}{\delta t} = \Delta \vec{p}$$
In other words, why does the rate of change of momentum equal to the total change in momentum?
If there were previous answers on this, please inform me! Thanks :)
A constant net force means:
$$\Sigma\vec{F}=\frac{d\vec{p}}{dt}=C$$
where $C$ is some constant. This means that
$$\int \ dp=p=C\int\ dt=Ct+p_0$$
where $p_0$ is the initial momentum. Now, you can easily verify that
$$p_2-p_1=\Delta p=Ct_2+p_0-Ct_1-p_0=C(t_2-t_1)=C\Delta t$$
In particular, you see that $\Delta p \neq \frac{dp}{dt}$, unless $\Sigma \vec{F}=0$, since then both equal zero.
I think I have figured out the answer, hopefully.
Firstly, let us begin by stating the First Law:
$$\sum \vec{F} = \frac{\delta \vec{p}}{\delta t}$$
When the net force is constant, we it means that there is no change of momentum, in other words,
$$ \frac{\delta \vec{p}}{\delta t} = 0$$
In this case, we know that the function of momentum is a constant, i.e, a straight line function.
This would derive the impulse-momentum theorem:
$$\Delta \vec{p} = \vec{J}$$
Would this be valid?
