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?

Answer 1

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.