Absolute Acceleration in a Train How would you calculate the total acceleration of a human body that is just starting to walk in the train against the direction of the train, who is also accelerating? Actually I want to have an evidence that the possibility to fall in a train while it is accelerating is higher, if you walk against the direction of the train than if you just stand.
 A: This is because of greater shear strain you will experience if you   run on the train while its accelerating. But us humans do not deform when subjected to this stress but they fall down.

There are two factors involved here

*

*Shear force(assuming uniform acceleration)

Without using complicated terms, let's assume that you're a  passenger(A) watching another passenger(A) standing inside the train. According to you when the train accelerates at $a *m/s^2$., the passenger(A) inside accelerates at $a *m/s^2$. Let his mass be $m*kg$. Humans generally run at an angle of $15 \deg$
Now person (A) has started to accelerate in the opposite direction at $a_1 *m/s^2$ at an angle of $15\deg$.Due to Newton's Third Law the force exerted by the train will oppose passenger A
Force at your feet(when running)
$$ \sum  F_{xz}=ma - \cos(15 \deg)ma_1$$
$$ \sum  F_{xz}=ma -  0.96ma_1$$
In very few cases this force would be in direction of $\overrightarrow{ma}$. Believe it or not except high speed bullets, normal trains accelerate very slowly.
Also due to Newton's Third Law, the forward acceleration of your head will be equal to the backward acceleration
Shear stress tends to deform originally rectangular objects into parallelograms. The most general definition is that shear acts to change the angles in an object.
$$\tau_{xz} = \frac{F}{A_{xz}}$$

And if you also want to know shear strain
$$ \tan \gamma_{zx} = \frac{\Delta x}{z}$$
$$\gamma_{zx} = \tan^{-1} \Bigr ( \frac{\Delta x}{z} \Bigr)$$

*

*z is height of the person

*$\Delta x$ = ?
$$\sin(15 \deg) = \frac{\Delta x - s_{net}}{z}$$
$$\sin(15 \deg)z = \Delta x - s_{net}$$
$$0.25z + s_{net} =\Delta x$$

*Let $a-0.96a_1$ be $a_{net}$
$$s_{net} = ut + \frac{1}{2}a_{net}t^2$$
Since humans are not rectangular, this is just an approximation, also air resistance and friction do not come in this approximation.
Now if passenger (A) were just to stand, he would not exert any force on the train, nor would the train exert any force on him due to Newton's third law. Thus
$$\tau_{xz} = \frac{0}{A_{xz}}$$
$$=0N/m^2$$
Since there is no force there is no displacement and person is perpendicular to the ground while standing,shear strain equals zero too.
$$\sin(90 \deg) = \frac{\Delta x -0}{z}$$
$$\frac{\Delta x -0}{z} = 0$$
$$\gamma_{xz} = \arctan (0)=0$$


*Friction
All of us know that without friction we won't be able to stand. While running friction  is lesser as compared to state of rest of an object in contact with the ground. Therefore a moving object will experience less friction, and will be more likely to fall.
Conclusion : You are more likely to fall if you run in the opposite direction of the train's acceleration than standing still on the train
