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Finally, the water park at an amusement park has rides that are essentially this lab on a large scale. A 45 kg rider falls a total distance of 22.5 m for one of these, sliding down an incline of 31.5 degrees. If there were no friction, how long would the ride last (while they are in the tube?) How fast would they exit the tube in m/s?

User Karissa
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Final answer:

To determine the duration of a water park ride and the exit speed of a rider sliding down a frictionless incline, we use kinematic equations and the principles of energy conservation, taking into account the angle of the incline and the components of gravitational acceleration.

Step-by-step explanation:

To calculate the duration of the ride and the exit speed of the rider in a water park slide, we can apply the kinematic equations and the principles of energy conservation for an object moving along an inclined plane without friction.

Duration of the Ride

To determine how long the ride would last, we can use the kinematic equation for constant acceleration, which states:

s = ut + (1/2)at²

We can find the acceleration component along the incline by decomposing gravity's acceleration using the angle of the incline. Since gravity is the only force acting on the rider, the parallel component of acceleration a is given by g sin(\( \theta \)), where g is the acceleration due to gravity (9.8 m/s²) and \( \theta \) is the angle of the incline. The initial velocity u is 0 m/s because the rider starts from rest.

Exit Speed of the Rider

The exit speed can be calculated using the conservation of mechanical energy or another kinematic equation. Since we have the inclined distance and acceleration, we can use the following equation:

v² = u² + 2as

Let's calculate the time and exit speed using the given values:

  • s = 22.5 m (distance fallen)
  • \( \theta \) = 31.5° (angle of incline)
  • a = g sin(31.5°)

By substituting the known values into the equations, we can solve for the time t and final velocity v.

User Yamen Nassif
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