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:
- Initial velocity, v1 = 6.00 m/s
- Final velocity, v2 = 19.0 m/s
- 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.