Final answer:
The two hockey players will collide in approximately 11.63 seconds. The accelerating player will be moving at a velocity of approximately 8.61 m/s at the time of contact. The accelerating player will have traveled approximately 3.50 m, while the second player will have traveled the full distance of 50 m.
Step-by-step explanation:
To solve this problem, we can use the equations of motion. Let's first calculate the time it takes for the two hockey players to collide. We can use the following equation:
s = ut + (1/2)at^2
where s is the distance, u is the initial velocity, a is the acceleration, and t is the time. For the first player, the initial velocity is 0 m/s, the acceleration is 0.74 m/s^2, and the distance is 50 m. Solving for time, we get:
50 = (1/2)(0.74)(t^2)
t^2 = 135.14
t = 11.63 s
Therefore, it would take approximately 11.63 seconds for the two players to collide.
To calculate the velocity of the accelerating player at the time of contact, we can use the equation:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. For the first player, the initial velocity is 0 m/s, the acceleration is 0.74 m/s^2, and the time is 11.63 s. Solving for the final velocity, we get:
v = (0)(11.63) + (0.74)(11.63)
v = 8.61 m/s
Therefore, at the time of contact, the accelerating player would be moving approximately 8.61 m/s.
To calculate the distance traveled by each player, we can use the equation:
s = ut + (1/2)at^2
where s is the distance, u is the initial velocity, a is the acceleration, and t is the time. For the first player, the initial velocity is 0 m/s, the acceleration is 0.74 m/s^2, and the time is 11.63 s. Solving for the distance, we get:
s = (0)(11.63) + (1/2)(0.74)(11.63^2)
s = 3.50 m
Therefore, the accelerating player would have traveled approximately 3.50 m. For the second player, since they are already at their maximum speed of 6.2 m/s, they would travel the entire distance of 50 m.