Answer:Hockey & Physics: Catching Up

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A hockey player accelerates from rest at 3.8 m/s² after a 3-second delay to catch an opponent skating at a constant speed of 12 m/s. The first player takes approximately 6.315 seconds to catch up, covering a distance of about 75.77 meters during this time. The calculations involve using kinematic equations for both players to determine the time and distance. The equations for the distances traveled by both players are set equal to find the catch-up point. Understanding these physics concepts is essential for solving similar problems in sports dynamics.
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Hockey and Physics!

Homework Statement



So I am good at hockey, but just not the physics aspect:-p. Could someone please check my answers and provide feeback


A hockey player is standing on his skates on a frozen pond when an opposing
player skates by with the puck, moving with a constant speed of 12 m/s. After
3.0 s, the first player makes up his mind to chase his opponent and starts
accelerating uniformly at 3.8 m/s^2.
(a) How long does it take him to catch his
opponent?
(b) How far does he travel before he catches up with his opponent?

Homework Equations





The Attempt at a Solution




Opposing skater is moving with constant velocity, a = 0 , v= 12m/s

After the 1st player starts and moves with an accleration of 3.8 m/s^2 after 3 seconds.

v= u+at = 3.8x3 = 11.4 m/s

Let "t" be the time in which the first player catches his opponent.

Within this time, the opponent moves a distance of s= ut= 12t m

therefore, the 1st player-> s= 1/2 a t^2 = 1.9t^2

12= 1.9t
t= 6.315 seconds


b) the distance traveled during this time period to catch his opponent is:

s= 1.9t^2= 1.9 (6.315)^2 = 75.77 m
 
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What is the equation for the first player?

x = 3*12 + 12*t

What is the equation for the guy starting out?

x = 1/2*a*t2

When the x is the same that will be the time.
 
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