# 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 his skates on a frozen pond when an opposing...

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 his skates on a frozen pond when an opposing player skates by the puck, moving with a constant speed of 12m/s. After 3.0 s, the first player makes up his mind to chase his opponent and starts accelerating uniformly at 3.8m/s^2.

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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.