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## 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. M

Question

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?

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Physics
3 years
2021-07-29T19:35:31+00:00
2021-07-29T19:35:31+00:00 1 Answers
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## Answers ( )

Answer: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:

RattlesnakeWhere:

– 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.

RoadrunnerWhere:

– 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.