A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 4.0 m/s, skates by with the puck. After 2.80 s, the first player makes up his mind to chase his opponent. If he accelerates uniformly at 0.26 m/s2, determine each of the following.
(a) How long does it take him to catch his opponent? (Assume the player with the puck remains in motion at constant speed.)

To solve this problem, we can approach it step by step. Let's define the variables first:

- Speed of the opposing player with the puck: \( v_o = 4.0 \, \text{m/s} \)
- Time delay before the first player starts chasing: \( t_d = 2.80 \, \text{s} \)
- Acceleration of the first player: \( a = 0.26 \, \text{m/s}^2 \)
- Time taken by the first player to catch up to the opponent: \( t \)

### Step 1: Determine the distance the opponent covers during the delay.

During the delay \( t_d \), the opponent is moving at a constant speed:
\[
d_o = v_o \times t_d = 4.0 \, \text{m/s} \times 2.80 \, \text{s} = 11.2 \, \text{m}
\]

### Step 2: Write the equations of motion.

1. **Distance covered by the first player after starting to chase:**

   The first player starts from rest and accelerates uniformly, so the distance covered after he starts chasing is given by:
   \[
   d_1 = \frac{1}{2} a t^2
   \]
   
2. **Distance covered by the opponent after the first player starts chasing:**

   The opponent continues to move at constant speed, so the distance covered by the opponent during this time is:
   \[
   d_2 = v_o \times t
   \]

### Step 3: Set up the equation for catching up.

For the first player to catch the opponent, the distance he covers \( d_1 \) must equal the distance the opponent has moved plus the distance covered during the delay:
\[
\frac{1}{2} a t^2 = d_o + v_o \times t
\]
Substitute the known values:
\[
\frac{1}{2} \times 0.26 \times t^2 = 11.2 + 4.0 \times t
\]
Simplify the equation:
\[
0.13 t^2 - 4.0 t - 11.2 = 0
\]

### Step 4: Solve the quadratic equation.

We solve this quadratic equation using the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 0.13 \), \( b = -4.0 \), and \( c = -11.2 \).

\[
t = \frac{-(-4.0) \pm \sqrt{(-4.0)^2 - 4(0.13)(-11.2)}}{2(0.13)}
\]
\[
t = \frac{4.0 \pm \sqrt{16 + 5.824}}{0.26}
\]
\[
t = \frac{4.0 \pm \sqrt{21.824}}{0.26}
\]
\[
t = \frac{4.0 \pm 4.672}{0.26}
\]

This gives us two possible solutions:
\[
t_1 = \frac{4.0 + 4.672}{0.26} \approx \frac{8.672}{0.26} \approx 33.35 \, \text{s}
\]
\[
t_2 = \frac{4.0 - 4.672}{0.26} \approx \frac{-0.672}{0.26} \approx -2.58 \, \text{s} \quad (\text{not physically possible})
\]

Since the negative time is not possible, the time it takes for the first player to catch his opponent is approximately \( t = 33.35 \) seconds.

### Final Answer:
(a) The first player takes approximately **33.35 seconds** to catch his opponent.

Would you like more details or have any questions about the solution? 

Here are some related questions you might find useful:

1. What is the total distance the first player covers before catching up?
2. How fast is the first player moving when he catches up to his opponent?
3. What is the relative velocity between the two players when the first player catches up?
4. If the first player started chasing after 5 seconds instead of 2.8, how would the result change?
5. What if the acceleration of the first player was higher, say 0.5 m/s²? How would that affect the time to catch up?

**Tip:** In problems involving relative motion and constant acceleration, breaking down the motion into parts and using the kinematic equations systematically can make complex problems easier to solve.

Explore how to solve a kinematics problem involving uniform acceleration and catching up with a moving opponent. This problem covers concepts such as equations of motion with constant acceleration and quadratic equations. Discover the step-by-step solution and learn how to apply these principles effectively. Suitable for high school students.

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