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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Vodka 2 months 2021-07-29T19:35:31+00:00 1 Answers 2 views 0

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    2021-07-29T19:37:07+00:00

    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 (v_{S} = 0.75\,\frac{m}{s}), whereas the roadrunner accelerates uniformly from rest. (v_{o, R} = 0\,\frac{m}{s}, a = 1\,\frac{m}{s^{2}})

    2) Initial distance between the roadrunner and rattlesnake is 10 meters. (x_{o, R} = 0\,m, x_{o,S} = 10\,m)

    3) The roadrunner catches up to the snake at the end. (x_{S} = x_{R})

    Now we construct kinematic expression for each animal:

    Rattlesnake

    x_{S} = x_{o,S}+v_{S}\cdot t

    Where:

    x_{o, S} – Initial position of the rattlesnake, measured in meters.

    x_{S} – Final position of the rattlesnake, measured in meters.

    v_{S} – Speed of the rattlesnake, measured in meters per second.

    t – Time, measured in seconds.

    Roadrunner

    x_{R} = x_{o,R} +v_{o,R}\cdot t +\frac{1}{2}\cdot a\cdot t^{2}

    Where:

    x_{o, R} – Initial position of the roadrunner, measured in meters.

    x_{R} – Final position of the roadrunner, measured in meters.

    v_{o,R} – Initial speed of the roadrunner, measured in meters per second.

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

    t – Time, measured in seconds.

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

    x_{o,S}+v_{S}\cdot t = x_{o,R}+v_{o,R}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}

    \frac{1}{2}\cdot a \cdot t^{2} + (v_{o,R}-v_{S})\cdot t + (x_{o,R}-x_{o,S}) = 0

    If we know that a = 1\,\frac{m}{s^{2}}, v_{o, R} = 0\,\frac{m}{s}, v_{S} = 0.75\,\frac{m}{s}, x_{o, R} = 0\,m and x_{o,S} = 10\,m, the resulting expression is:

    0.5\cdot t^{2}-0.75\cdot t -10=0

    We can find its root via Quadratic Formula:

    t_{1,2} = \frac{-(-0.75)\pm \sqrt{(-0.75)^{2}-4\cdot (0.5)\cdot (-10)}}{2\cdot (0.5)}

    t_{1,2} = \frac{3}{4}\pm \frac{\sqrt{329}}{4}

    Roots are t_{1} \approx 5.285\,s and t_{2}\approx -3.785\,s, 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.

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