A roadrunner at rest suddenly spots a rattlesnake slithering directly away at a constant speed of 0.75 m/s. At the moment the snake is 10. Meters from the bird, the roadrunner starts chasing it with a constant acceleration of 1.0 m/s^2. How long will it take the roadrunner to catch up to the snake?

The roadrunner will take approximately 5.285 seconds to catch up to the rattlesnake.

Explanation:

From the statement we notice that:

1) Rattlesnake moves a constant speed (), whereas the roadrunner accelerates uniformly from rest. (, )

2) Initial distance between the roadrunner and rattlesnake is 10 meters. (, )

3) The roadrunner catches up to the snake at the end. ()

Now we construct kinematic expression for each animal:

Rattlesnake

Where:

- Initial position of the rattlesnake, measured in meters.

- Final position of the rattlesnake, measured in meters.

- Speed of the rattlesnake, measured in meters per second.

- Time, measured in seconds.

Roadrunner

Where:

- Initial position of the roadrunner, measured in meters.

- Final position of the roadrunner, measured in meters.

- Initial speed of the roadrunner, measured in meters per second.

- Acceleration of the roadrunner, measured in meters per square second.

- Time, measured in seconds.

By eliminating the final positions of both creatures, we get the resulting quadratic function:

If we know that , , , and , the resulting expression is:

We can find its root via Quadratic Formula:

Roots are and , respectively. Both are valid mathematically, but only the first one is valid physically. Hence, the roadrunner will take approximately 5.285 seconds to catch up to the rattlesnake.

i wanna say it’s between b and d, leaning more towards b.

0.19 m/s and 0.45 m/s