# Acceleration of a car in a collision?

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?

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.

• Right. More specifically, the car has a momentum, a quantity which is always conserved :) Dec 4, 2013 at 2:24

Another way to think about Newton's second law (and the way he originally defined it) is $F=\dfrac{d\rho}{dt}$, where $\rho=mv$ is momentum and $\dfrac{d\rho}{dt}$ is the rate of change of momentum. I think you meant to say that the obstacle will exert a force on you - and that is correct. If you could calculate your change in velocity, and the amount of time you were in contact with the obstacle, you could use impulse $I=\Delta \rho = F\cdot t$ to find the (average) force the obstacle exerts on your car during the collision.

When you hit the obstacle (if you don't destroy it) your speed goes down to zero in a quite small time. That gives you the acceleration. It starts when you start hitting and it ends when you come to a complete stop. This acceleration is due to the force that the obstacle generates on the car.

Think about: $\displaystyle a=\frac{\Delta v}{\Delta t}=\frac{v_i}{t_\text{hit}}$.

The bigger is the initial velocity, the bigger will be the acceleration. If your car can progressively deform itself, then you extend the hitting time and you reduce the acceleration involved, that's what cars constructors always try to do.

When I read about this question, I was thinking, well, "a" can't be zero because it should represent your deceleration from 900 miles/hour. The resulting F would be the force you experience when you decelerate. (The deceleration depends on how you hit the obstacle, how long did it take you to stop/slow, across what distance, and the degree of deformity of your vehicle.)