Example: If a car with mass $m=1000 kg$ and with velocity $v=30m/s$ crashes into a wall (say without applying the breaks) and comes to rest, it is easy to calculate the total change in momentum
$$\Delta p~=~0-mv~=~\int F(t) dt,$$
also known as the impulse, but it is difficult to know the precise force $F(t)$ on the car as a function of time $t$ during the impact as the front of the car (and the wall) folds up in some particular way.
Unless I made some mistakes, impulse is equal to momentum and not to
change of momentum. Where did I go wrong, or, what is the final word?
I'm not sure I completely understand your notation in your reasoning preceding the statements quoted above but 3rd line certainly doesn't look correct:
Acceleration = a=[1]Δv change of velocity (m) v/s
The correct equation is
$$\bar a = \frac{\Delta v}{\Delta t} $$
where $\bar a$ is the average acceleration. With this correction, the final equation becomes
$$J = (m)\bar a \Delta t = (m) \Delta v = \Delta p$$
This seems so straightforward that I suspect I don't understand what you're actually trying to show.
what is change of velocity then? if a football is at rest and I kick
it and it aquires v=10m/s, haven't I accelerated it over a period of
time? isn't that difference of Δv=+10m/s acceleration?
Yes, you accelerated it over a period of time and no, the difference in velocity is not acceleration.
(Average) acceleration is, as I wrote above, the ratio of the change in velocity to the elapsed time over which the change occurred.
So, if the change in velocity is
$$\Delta v = 10 \frac{m}{s}$$
one does not know the average acceleration unless one also knows the elapsed time $\Delta t$ since the average acceleration is given by
$$\bar a = \frac{\Delta v}{\Delta t}$$
Clearly, the average acceleration is inversely proportional to the elapsed time so, the smaller the elapsed time, the larger the average acceleration.
To be concrete, let us say that the foot was in contact with the football for $100 ms$. Then, the average acceleration of the football is
$$\bar a = \frac{10 \frac{m}{s}}{0.1s} = 100 \frac{m}{s^2} $$
Best Answer
Just to clarify, impulse is force times time. Which one matters depends on what you are trying to achieve.
If you want to propell a ball thru the air, then it's the impulse that matters. Note that a ball whacked for a short time by a bat at high force can go further than you can throw by applying lower force and much longer time.
If you are trying to break a board with a karate chop, it is largely the force that matters. Once the board has cracked, which happens quickly due to the shock wave, additional applied force is of little benefit.