accelerationcollisionforcesnewtonian-mechanics
For example because of newton law $F=ma$ force is acceleration times mass but if I go at a high constant velocity ($900\: \mathrm{miles/hour}$) in my car for examples I have $0$ acceleration but if I hit an obstacle by newton's law it will have no effect on me because $F=ma$ and mass of my car x 0 = 0
But if I do the experiment I will find that I actually hit the obstacle and it does an force on me why?
I stumbled on a reference which summarizes what I think is the most correct answer to this question. You have several options, but I think we should only use terms that we have a physical reason to write. Here's the reference:
http://usna.edu/Users/physics/schneide/Buick.htm
They use a lot of unnecessary details like time between stopping that we're not interested in. I agree with their graph but not their equation. So here is my equation to explain their graph. I also limited it to flat roads (no hills).
$$F = \frac{P_0}{v} + \mu mg + c \rho A \frac{v^2}{2} $$
From the reference, known values about their car are:
I report these because there is no direct measurement available. I would then use their data to evaluate the three coefficients that determine the relative friction from each thing.
They put a constant term in the transmission factor (the 1/v term). I don't like this, because I want clean mathematics, so I'm bunching that long tail of transmission with the friction coefficient.
$$ \mu (1800 kg) (9.8 m/s^2) = 200 N + 250 N = 450 N \rightarrow \mu = 0.026 $$
$$ \frac{P_0}{15 m/s} = 500 N - 250 N = 250 N \rightarrow P_0 = 3750 W$$
$$ c (3 m^2) (1.3 kg/m^3) \frac{(31 m/s)^2}{2} = 500 N \rightarrow c = 0.27 $$
These are all consistent with what the link claimed, aside from the cases where I willfully used a different kind of definition. Another good thing that these all have physical interpretations, which those units are suggesting. I will avoid getting into the exact interpretation because I feel like there's space for quibbling.
I plotted this on Wolfram alpha. This is my altered version of that link.
This fits expectations fairly well. Take note, however, of the 1/v term. That represents fuel consumption due to constant loads (thus, units of power, of course). That might not be relevant if you're looking for a force, but it can be kind of interpreted as a force. It's a force the engine is exerting against itself (to some fraction of that number) due to idling. It is also the constant electronic loads on the battery... and the charging of the battery itself. It's not the friction of the wheels on the road of air on the car. If you're only interested in those then you might do well to just take the last two terms. If you do that, however, there is no concept of maximum gas mileage, nor should there be. What you're going to do with these terms depends on the application. I just believe this to be the best available option so I posted it.
The $F$ in the equation $F=ma$ is not the force that would be exerted by the object if it were to hit something else. Instead, $F$ represents the net force acting on the object that must be present in order to produce the current acceleration $a$ of the object. A better way to write Newton's second law is $$F_\text{net}=ma,$$ since it shows explicitly which force is being represented on LHS of the equation is.
In your train example, if the train is traveling at a constant velocity of 100 mph, the acceleration is zero, and by Newton's second law the net force is also zero. But this has no bearing on what force the train would exert on something if it collided.
Best Answer
You're confusing the acceleration of your car with the acceleration in a collision. You actually have to look at it "backwards" from what you've described above.
That is, in the collision you don't do a $F = ma$ calculation where $a$ is the acceleration of your gas pedal. Instead in the collision you have a force $F$ resulting from the collision and you do $a = F/m$ to get the resulting acceleration your car is subjected to.