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In a drill during basketball practice, a player runs the length of the 30-meter court and back. The player does this three times in 60 seconds. The magnitude of the player's average velocity during the drill is...
1. 0 m/s
2. 0.5 m/s
3. 1 m/s
4. 1.5 m/s
5. 3 m/s
6. 30 m/s

Respuesta :

Answer:

3.0m/s so it’s number 5

Explanation:

onyebuchinnaji

The magnitude of the player's average velocity during the drill is 0 m/s.

The given parameters;

  • length of the court, L = 30 m
  • number of times, = 3
  • time of motion, t = 60 s

Average velocity is change in displacement per change in time of motion.

[tex]\Delta \bar v = \frac{\Delta x}{\Delta t}[/tex]

where;

  • [tex]\Delta x[/tex] is the change in displacement

The change displacement is zero, since the player moved forward and back to the exact initial position.

[tex]\Delta \bar v = \frac{3(30) - 3(30) }{60} = 0[/tex]

Thus, the magnitude of the player's average velocity during the drill is 0 m/s.

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