How long does it take him to catch his opponent and how far does he travel before he catches up with his opponent? A hockey player is standing on...

The opposing player has a constant speed of  12m/s. Since
the first player starts after 3seconds, the second player is already 12m/s*3s = 36 m
away. Thus the initially the second player is 36 meters 
ahead.

Now let us say after  t seconds the 1st player
catches the  second player. So the 2nd player moves  in t seconds by a distance = speed
* time = 12t meters.

The 1st player has a starting speed =
0 m/s and ending speed  after a time of t seconds = acceleration* t = 3.8*t m/s
.

So the average speed of the 1st player = (0+3.8t)/2 =
1.9t m/s.

So the distance covered by the 1st player in time
t seconds = his average speed* time t seconds = (1.9t)*t = 1.9t^2 . But this distance
should be equal to 36 meters of initial distance  + 12t meters covered in t  seconds by
the 2nd plater. Therefore the rquired eqution is :

1.9t^2 =
36+12t. This is a quadratic equation in t.

We multiply the
equation by 10 to get integral coefficients:

19t^2 =
360+120t.

We rearrange the equation making zero on the
right.

19t^2 -120t -360 =
0.

Using quadratic formula, we get t = {-(-120) +sqrt(120^2
- 4*19*(-360)}/(2*19) = 8.5356secons.

The distance
travelled by the first player = 1.9t^2 =  138.4271
m.

Therefore the time after which the 1st player catches
the 2nd player = 8.5356 seconds.

The distance the first
player moves to catch the second player = 138.4271 m.