A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 14.3 m/s, skates by with the puck. After 2.2 s, the first player makes up his mind to chase his opponent. If he accelerates uniformly at 4.6 m/s2, how long does it take him to catch his opponent?
Displacement, Velocity and Acceleration
In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.
Linear Displacement
The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.
![### Problem Statement
A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 14.3 m/s, skates by with the puck. After 2.0 seconds, the first player makes up his mind to chase his opponent. If he accelerates uniformly at 4.6 m/s², how long does it take him to catch his opponent?
#### Instructions:
Round your answer to 1 decimal place.
### Explanation:
In this problem, we need to determine the time it takes for a hockey player, starting from rest and with a uniform acceleration, to catch up with another player moving at a constant speed.
1. **Identify Known Variables:**
- Speed of opposing player (v₁): 14.3 m/s
- Time before chase starts (t₀): 2.0 s
- Acceleration of first player (a): 4.6 m/s²
2. **Calculate Distance Covered by Opposing Player During Initial Time:**
- Distance traveled by the opposing player in t₀ seconds = speed × time = 14.3 m/s × 2.0 s = 28.6 meters
3. **Setup Equations for Positions:**
- Distance covered by opposing player after t seconds (d₁): d₁ = 14.3 m/s × (t + 2.0 s)
- Distance covered by first player after t seconds (d₂): d₂ = 0.5 × 4.6 m/s² × t²
4. **Determine the Time to Catch Up:**
- Set d₁ equal to d₂ and solve for t
- 14.3(t + 2.0) = 0.5 × 4.6 × t²
- Simplify and solve the quadratic equation to find t.
5. **Round to 1 Decimal Place:**
### Questions to Consider:
- How do changes in initial velocities affect the time to catch up?
- What roles do acceleration and time delay play in this scenario?
### Applications:
This problem highlights key physics concepts such as uniform motion, uniform acceleration, and quadratic equations, useful in various fields including sports physics, automotive design, and any context where predictive modeling of moving objects is relevant.
### Diagram (if applicable):
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