For example, let's say a car, weighing 1300kg (m=1300kg) is on cruise control, going at a constant 80km/h. That essentially means that there is no acceleration or deceleration, as the motion is uniform. a=0. Now, let's say the car didn't see where it was going and it hits the side of a building.

Since F = ma:

F = 1300 * 0

F = 0 Newtons

This doesn't make sense to me. There is no way that the force could be 0. Could someone explain this to me?

  • $\begingroup$ Do you agree that the the car's velocity vector probably changed as a result of the impact..? $\endgroup$ – user76568 Oct 10 '14 at 13:02
  • $\begingroup$ Yes, it makes sense that if the car suddenly changes from 80 to 0 in a short time, it counts as a (negative) acceleration. Thanks! $\endgroup$ – Admin Voter Oct 10 '14 at 13:14
  • $\begingroup$ It's like this: IF you apply a force, it will accelerate the object by a certain amount. That "amount" is given by the formula. $\endgroup$ – Fattie Oct 10 '14 at 16:37

If the car just rolls at constant speed as you say, then you are right, there is no acceleration at all:

$$a=0 \Rightarrow$$ $$F=m a=0 $$

which means that there is no net force acting on the car at all (Newtons first law of motion). It just continues without disturbance.

But if the car hits the wall... Then there is an acceleration! Actually a deceleration if you wish to call it that (that is, negative acceleration $a<0$). And that acceleration must be (numerically) enormous (let's write it as $a<<0$), since the car is being stopped suddenly by the wall and forced from it's initial speed down to zero speed in a very short time (the time the collision lasts - from the car touches the wall 'till it has been fully stopped - is extremely short).

Then you have (Newtons second law of motion):

$$a << 0 \Rightarrow$$ $$F=m a<<0 $$ and the force that is done on the car to cause this sudden stop is (numerically) extremely large. And the car is whacked.

  • $\begingroup$ Ah yes, I didn't realize that, at the time of impact, the speed actually changes from 80 to 0, which technically counts as a deceleration! So would the time be around a few milliseconds in that case? Edit: by the way, very thorough answer, thanks. $\endgroup$ – Admin Voter Oct 10 '14 at 13:11
  • $\begingroup$ As far as I remember the time $t$ in car collisions is usually in milliseconds, yes. It depends on the speed combined with how the front of the car behaves. A trick to easen the effect of a collision to improve safety is namely to design the front to "crumple" as much as possible before a full stop. Then the change in speed happens more gradually and the (numerical) acceleration is smaller. $\endgroup$ – Steeven Oct 10 '14 at 13:19
  • $\begingroup$ Literally during a car crash, imagine a test vehicle hitting a VERY hard wall (imagine a steel wall which is "totally immovable"). Then the very front tip of the car experiences incredible acceleration. It takes "no time" for the very front of the car to "totally stop." As the remainder of the car crumples the other parts of the car take more time, and hence experience less acceleration (less force). This is why, thanks to the utterly amazing engineering of modern cars, you can sometimes survive a crash. $\endgroup$ – Fattie Oct 10 '14 at 16:40

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