Answer:Hockey & Physics: Catching Up

[cbrowne]

Hockey and Physics!

Homework Statement

So I am good at hockey, but just not the physics aspect. 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 acceleration 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

 

[LowlyPion]

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.