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Elenna [48]
3 years ago
9

A rugby players runs 10.0 m straight down the playing field in 2.00 s. The player is then hit and pushed 5.00 m straight backwar

d in 1.50 s. The player breaks the tackle and runs straight forward another 17.2 m in 4.10 s. Calculate the velocity for the players entire motion.
Physics
1 answer:
s344n2d4d5 [400]3 years ago
3 0

Answer:

2.92 m/s

Explanation:

For the entire motion, the average velocity (V) can be calculated thus;

V = \frac{displacement}{time}

Total displacement of the player = 10.0 + (-5) + 17.2

        = 22.2 m

Total time = 2.00 + 1.50 + 4.10

         = 7.60 s

V = 22.2/7.60

       = 2.92 m/s

Hence, the velocity of the players for the entire motion is 2.92 m/s.

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v₂ = The speed of the student just before she lands

ω₂ = The angular velocity of the wheel just before she lands

By the conservation of energy, we have;

P.E.₁ + K₁ = P.E.₂ + K₂

∴ m·g·h₁ + \dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2 = m·g·h₂ + \dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2

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ω₂ = v₂/r

∴ 5061.96 J + 0 = 0 + \dfrac{1}{2} \times 43.0 \, kg \times v_{2}^2+\dfrac{1}{2} \times 9.60 \, kg\cdot m^2 \cdot \left (\dfrac{v_2}{0.300 \, m} }\right ) ^2

5,061.96 J = 21.5 kg × v₂² + 53.\overline 3 kg × v₂² = 21.5 kg × v₂² + 160/3 kg × v₂²

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v₂ ≈ √(67.643118 m²/s²) ≈ 8.22454363 m/s

The speed of the student just before she lands, v₂ ≈ 8.225 m/s.

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