Good ideas that don't work so well

We work in the non-inertial frame of the train. Since the train is accelerating forwards, the man experiences a fictitious force backwards (opposite the frame's direction of acceleration). The magnitude of this force is $F_{fict}=ma_{train}$ where $m$ is the man's mass and $a_{train}=2m/s^2$ is the magnitude of the train's acceleration.

Friction acts forwards to counteract this fictitious force, and its magnitude is given by $f=\mu mg$ . This gives the net force as $F_{net}=ma=F_{fict}-f$ . Dividing throughout by $m$ gives $a=1.02m/s^2$ backwards.

Since the man starts at the midpoint of the car, he will travel 15 meters before hitting the backwall. Using $s=1/2 at^2$ and solving for $t$ gives 5.42 seconds.

In this problem, the car is a non-inertial frame of reference, due to the fact that it is accelerating. The rich guy cannot in turn write Newton's 2nd law in the usual format, but instead he must subtract an $ma$ term, where $a$ the acceleration of the car, from the vector sum of all the forces. Hence his acceleration is given by:

$f-ma=ma'$

$f=μN$

$N=mg$

which imply that $a'=a-μg=1.02 m/s^2$ .

By integrating, we find that $x_{rich}=\frac{1}{2}a't^{2}\Rightarrow{t=\sqrt{\frac{2x}{a'}}}$

and after the substitution $x_{rich}=15 m$ , $a'=1.02 m/s^2$ we obtain $t=5.423 sec$ . μ is the coefficient of friction.

We're given the following information:

• There is a constant acceleration towards the back of the train car, namely $a = 2 \frac {m}{s^2}$ .
• There is a resisting force due to friction towards the front of the train $F_{friction} = \mu N$ . Where $\mu_s = \mu_k = \mu = 0.1$ and $N = -mg = 9.8*m$ $Newtons$
• The man is standing in the middle of a 30 meter car. So, $x_o = 15$ $meters$ if we assume that the front of the car is at $x_o=0$ and the back is $x_o=30$ .

With the above information we can calculate the total force acting on the man towards the back of the car.

$F_{total} = ma - \mu N$

$= m(2-0.1*9.8)$

$= m*1.02$

Since $F = ma$ we know that the net acceleration acting on the man towards the back of the car is $1.02 \frac {m}{s^2}$ . The equation for position $x$ at time $t$ is

$x(t)=\frac {1}{2} at^2+v_ot+x_o$ .

Solving for x(t)=30 gives

$30 = \frac {1}{2} 1.02t^2+0v_o+15$

$30 = 0.51t^2+15$

$t = \sqrt{\frac {15}{0.51}} = 5.42$ $seconds$

The acceleration of the rich guy is provided by the frictional force acting on him, which is $F_f=\mu mg$ . The maximum acceleration of the train without him slipping is therefore $\mu g=0.98$ . Hence the rich guy will start to slip. If we choose coordinates so that the left edge of the train at t=0 is at x=0, then the rich guy's position as a function of time is given by = $x_g=15+\frac{1}{2} 0.98 t^2$ . Similarly, the position of the left edge of the train car is $x_t=\frac {1} {2} 2 t^2$ . Setting these two equal to each other yields t=5.423 seconds.

pseudo accelaration =2 si units.frictional accelaration(opposite to pseudo accelaration)=neu g=0.1 9.8=0.98 Net accelararion=pseudo accelaration-frictional accelaration =2-0.98=1.02 si units Now use second equation of motion n put u=0, a=1.02 and S=15, to calculate t, which then comes out to be 5.432 sec.

Since the train is accelerating, the man is standing in a non-inertial reference frame, and thus experiences a fictitious force in the opposite direction. If the floor was frictionless, he would accelerate towards the back of the car at a=2m/s^2. However, since there is a force of friction, his net acceleration would be a-μg=1.02m/s^2. Since he has to travel a distance d=15m(he is at the middle of the car), and starts at rest, the time taken is sqrt(2d/a)=sqrt(2*15/1.02) which is around 5.423s.

Force = mass x acceleration (F=ma) - we don't know the man's mass for this, but it doesn't matter. The key thing is that the train is moving underneath him, so he is moving relative to the train.

Draw a diagram with a box representing the man on a level surface representing the train carriage. Four arrows: - one pointing down for the force created by gravity at 9.8m - one pointing up, perpendicular to the floor, for the normal reaction(R) (also 9.8m because he's in contact with the ground on a level surface and not going upwards) - one pointing left for the force created by the acceleration of the train - it's actually moving to the right, but everything's relative (2m) - one pointing right for friction, which is coefficient of friction (µ) x the normal reaction (R), or 0.1 x 9.8m = 0.98m.

The net force on the man is 2m - 0.98m = 1.02m, and since F=ma, the net acceleration is 1.02m/s.

Substitute that in to s=ut+0.5at^2 where s = distance = 15m and u = initial speed = 0, and you get:

15 = 0t+0.5(1.02)t^2 15 = 0.51t^2 t^2 = 29.41, and since t must be positive t = 5.423