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cbrowne
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Homework Statement
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
We know the initial velocities and accelerations of the two players,
a1 = 4 m/s2, a2 = 0
v1i = 0, v2i = 12 m/s
x1i = x2i = 0
The position of player #2 is given by
x2 = x2i + v2i t + (1/2) a2 t2
x2 = (12 m/s) t
Be careful with the time. We must account for player #1's wait of 3 s. With this accounted for, we can calculate the position of player #1 from
x1 = x1i + v1i (t - 3 s) + (1/2) a1 (t - 3 s)2
Of course, this equation only makes sense for t > 3 s.
x1 = 0 + 0 + (1/2) (3.8 m/s2) (t - 3 s)2 = (2 m/s2) (t2 - 6 s t + 9 s2)
Now we set x1 = x2 and solve for the time t.
x1 = x2
(1.9 m/s2) (t2 - 6 s t + 9 s2) = (12 m/s) t
We can either keep the units in explicitly or ensure that we have consistent units and simply write
1.9 (t2 - 6t + 9) = 12t
1.9 t2 - 12 t + 18 = 12 t
1.9 t2 - 24 t + 18 = 0
There are two solutions to this quadratic equation,
t1 = 11.8 s, and t2 = 0.8 s
However, the equation for the position of player #1 is not valid for t2 < 3 s, so we keep only tx,
t = 11.8 s
(b) How far has the first player traveled in this time?
Now where is player #2 (and, therefore, player #1 as well) at this time?
x2 = x2i + v2i t + (1/2) a2 t2
x2 = (12 m/s) t
x2 = (12 m/s) (11.8 s)
x2 = 141.6 m