Let v_s and v_r represent the velocities of the swimmer and river respectively with respect to ground
v_s + v_r = 12  ............(1)
and   v_s - v_r = 6  ............ (2) 
Adding the both equations (1) and (2) 2v_s = 12 + 6 = 18 km h^-1 or v_s = 9 km h^-1
From equation (1),
9+ v_r = 12 or v_r = 3 km h^-1
When the river flow and swimmer move in the same direction, the net velocity of swimmer is 12 km h^-1.

Relative velocity of B with respect to A
 \({ v }_{ BA }={ v }_{ B }-{ v }_{ A }\)
= 50 km h^-1 + (-50) km h^-1 
= 0 kmh^-1
Similarly, relative velocity of A with respect to B i.e., v_AB is also zero.
Thus each train will appear to be at rest with respect to the other.

The final velocity of the particle v = 0
The initial velocity of the particle
\(u=54\times\frac{5}{18}\ ms^{-1}=15\ ms^{-1}\)
S = 225 m
Retardation is always against the velocity of the particle.
v² = u² - 2aS
0 = (15)² - 2a(225)
450 a = 225
\(a=\frac{225}{450}\ ms^{-2}=-0.5ms^{-2}\)
Hence, retardation = 0.5 m s^-2.

\(\vec v=\frac{\bar {dr}}{dt}=6t\hat i+5\hat j\)
The velocity at any time 't' is given by \(\vec v=6t\hat i+5\hat j\) . The magnitude of velocity is speed.
The speed at any time 't' is then given by
\(speed=\sqrt {(6t)^2+5^2}=\sqrt {6t^2+25}\)
Now the velocity at t = 3 sec is given by
\(\vec v=6(3)\hat i+5\hat j=18\hat i+5\hat j\)
and speed at t = 3 sec, is given by
\(speed=\sqrt {349}\ m/s\)

The relative velocity of B with respect to A, \(\vec { { V }_{ BA } } =\vec { { V }_{ B } } -\vec { { V }_{ A } } \)= 5 km h^-1 due west similarly, the relative velocity of A with respect to B i.e., \(\vec { { V }_{ AB } } =\vec { { V }_{ A } } -\vec { { V }_{ B } } \) = 5 km h^-1. To a passenger in the car A, the car B will appear to be moving east with a velocity 5 km h^-1. To a passenger in train B, the train A will appear to move westwards with a velocity of 5 km h^-1