I try to find a solution for this Problem: We should compare circular motion with uniform acceleration.

The Task is to find out whether it is better to make a turn if you face a wall in front of your car or to brake.Turn would mean a 90° turns so that you do not hit the wall. Another important information is that the velocity which you are using to turn should equal the negative acceleration (brake).

What I tried to do: $u(circle)=2\pi r$ I face the wall with 90° so $S=r (circle)$ S is the distance until I hit the wall: $s(u)=vt $, $t=\frac {s(u)}{v} $, $s(u,90°)=\frac {\pi r}{2}$, $t=\frac {\pi r}{2v} $

However I need $r$ because this is the distance to the wall; $s=\frac {2vt}{\pi} $

Then I described the negative acceleration when $v=0\frac {m}{s} $:$s=\frac {V0^2}{2a} $ Now I wanted to check how the equations behave when $a$ respectively $v$. As I said before they both need to be the same; $\lim_{a \rightarrow \infty} \frac {V_0^2}{2a}=0$ and $\lim_{v \rightarrow \infty} \frac {2vt}{ \pi}=\infty$

One goes to 0 one to $\infty$. Is there any way to express where it does not matter to turn or to brake.

  • $\begingroup$ Why would it not always make more sense to brake? $\endgroup$ – Ben51 Feb 12 '18 at 19:38
  • $\begingroup$ When brakes are shot ( stopping distance is significantly greater than the turn radius)? $\endgroup$ – Charles Bretana Feb 12 '18 at 20:22

Like you have already found, you need an accelaration of $a=\frac {V_0^2}{2R}$ if you brake in a straight line and come to a stop right before the wall at distance R.

If you make a turn in the shape of a perfect circle, you need an accelaration of $a=\frac {V_0^2}{R}$, the centripetal accelaration.

So it is always better to brake in a straight line as that requires less accelaration.

  • $\begingroup$ Unless you can avoid the obstacle by turning. But yes, the shortest stopping distance is when all the grip goes towards removing speed, and not towards turning the car. $\endgroup$ – ja72 Feb 12 '18 at 21:03

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