Why do equation of motion fail to apply in non-inertial frame even after applying pseudo force?

Question

I considered two cars of equal mass moving towards each other with speed 30m/s and 18 m/s respectively and when they are 100 meters apart they both start de accelerating at 6m/s² each. So if we consider ground as reference and applying formula $v²- u²= 2as$ where $v$ and $u$ are initial and final velocity and a and s are distance and acceleration, then we get car 1 travelled 75m before stopping and car 2 travelled 27 m So total 102 m which means collision occur

However if we consider car 1 as frame of reference and consider a pseudo acceleration on car 2 of 6m/s² then here u=18+30ie 48 and v=0 and a= -12 So we get distance s by same formula as 96 which means no collision

So why did even after considering pseudo force do equation of motion fail?

Answer 1

Your computation in the inertial frame seems to be correct. However, car 2's deceleration in car 1's frame is wrong. Let's look at this in detail.

Let $D$ be the distance between the car when they start to slow down.

Let $F$ be the inertial frame (the road) and $F_1$ the frame that moves with the car.

Let $v_{10}$ and $v_{20}$ be the initial velocities of both cars in $F$.

The acceleration of car 2 in $F_1$ is:

$$a_2(F)=a_2(F_1)+a(F_1/F)\quad\Rightarrow\quad a_2(F_1)=-2a$$

The initial velocity of car 2 in $F_1$ is given by:

$$v_2(F)=v_2(F_1)+v(F_1/F)\quad\Rightarrow\quad v_{20}(F_1)=-v_{10}-v_{20}$$

Newton's Law for car 2 in $F_1$ is:

$$ a_2=-2a\\ v_2=-2at-(v_{10}+v_{20})\\ x_2=-at^2-(v_{10}+v_{20})t+D $$

Since car 2 will stop (in $F$) before car 1, let $t_2$ be the time when car 2 stops. Its value is the same in all frames since this is newtonian mechanics. At this instant, in $F_1$, car 2's velocity is:

$$v_2(F)=v_2(F_1)+v(F_1/F)\quad\Rightarrow\quad v_2(F_1)(t_2)=-v_1(F)(t_2)$$

So we have to compute car 1's velocity in $F$ at this instant. You already did this computation, so I simply give the result:

$$v_1(F)=-at+v_{10}\quad\Rightarrow\quad v_1(F)(t_2)=-at_2+v_{10}$$

Then, back in $F_1$:

$$ v_2(t_2)=-2at_2-(v_{10}+v_{20})=-(-at_2+v_{10})\\ t_2=\frac{v_{20}}{a} $$

This is the same result as in $F$, so the distance travelled by car 2 is the same in both frames. The distance travelled by car 1 must be computed in $F$ since it's at rest in $F_1$, so the result is unchanged.