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A cyclist competing in the tour de france coasts down a hill, dropping through a vertical distance of 30.0 m. the cyclist has an initial speed of 6.00 m/s and a final speed of 19.0 m/s.

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Final answer:

A cyclist competing in the tour de france coasts down a hill, dropping through a vertical distance of 30.0 m. The cyclist has an initial speed of 6.00 m/s and a final speed of 19.0 m/s. Using the equations of motion, we can find the acceleration of the cyclist and the time taken to travel the given distance.

Step-by-step explanation:

In this problem, we can use the equations of motion to find the final velocity and distance traveled by the cyclist.

Given:

  1. Initial velocity, v1 = 6.00 m/s
  2. Final velocity, v2 = 19.0 m/s
  3. Vertical distance, h = 30.0 m

Using the equation v22 = v12 + 2ah, where a is the acceleration and h is the vertical distance, we can solve for the acceleration:

a = (v22 - v12) / (2h)

Substituting the given values, we get:

a = (19.02 - 6.002) / (2 * 30.0)

a ≈ 3.90 m/s2

Next, we can use the equation s = v1t + (1/2)at2, where s is the distance traveled, t is the time, and a is the acceleration, to find the time taken to travel the given distance:

Substituting the values, we get:

30.0 = 6.00t + (1/2)(3.90)(t2)

Using the quadratic formula, we can solve for t:

t = (-b ± √(b2 - 4ac)) / (2a)

Here, a = (1/2)(3.90), b = 6.00, and c = -30.0. Plugging in these values, we find two solutions for t:

t ≈ 2.55 seconds or t ≈ -5.85 seconds

Since time cannot be negative in this context, we take t ≈ 2.55 seconds as the time taken to travel the given distance.

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