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:
Rattlesnake
![x_{S} = x_{o,S}+v_{S}\cdot t](https://tex.z-dn.net/?f=x_%7BS%7D%20%3D%20x_%7Bo%2CS%7D%2Bv_%7BS%7D%5Ccdot%20t)
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
![x_{R} = x_{o,R} +v_{o,R}\cdot t +\frac{1}{2}\cdot a\cdot t^{2}](https://tex.z-dn.net/?f=x_%7BR%7D%20%3D%20x_%7Bo%2CR%7D%20%2Bv_%7Bo%2CR%7D%5Ccdot%20t%20%2B%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20a%5Ccdot%20t%5E%7B2%7D)
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:
![x_{o,S}+v_{S}\cdot t = x_{o,R}+v_{o,R}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}](https://tex.z-dn.net/?f=x_%7Bo%2CS%7D%2Bv_%7BS%7D%5Ccdot%20t%20%3D%20x_%7Bo%2CR%7D%2Bv_%7Bo%2CR%7D%5Ccdot%20t%20%2B%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20a%20%5Ccdot%20t%5E%7B2%7D)
![\frac{1}{2}\cdot a \cdot t^{2} + (v_{o,R}-v_{S})\cdot t + (x_{o,R}-x_{o,S}) = 0](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20a%20%5Ccdot%20t%5E%7B2%7D%20%2B%20%28v_%7Bo%2CR%7D-v_%7BS%7D%29%5Ccdot%20t%20%2B%20%28x_%7Bo%2CR%7D-x_%7Bo%2CS%7D%29%20%3D%200)
If we know that
,
,
,
and
, the resulting expression is:
![0.5\cdot t^{2}-0.75\cdot t -10=0](https://tex.z-dn.net/?f=0.5%5Ccdot%20t%5E%7B2%7D-0.75%5Ccdot%20t%20-10%3D0)
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)}](https://tex.z-dn.net/?f=t_%7B1%2C2%7D%20%3D%20%5Cfrac%7B-%28-0.75%29%5Cpm%20%5Csqrt%7B%28-0.75%29%5E%7B2%7D-4%5Ccdot%20%280.5%29%5Ccdot%20%28-10%29%7D%7D%7B2%5Ccdot%20%280.5%29%7D)
![t_{1,2} = \frac{3}{4}\pm \frac{\sqrt{329}}{4}](https://tex.z-dn.net/?f=t_%7B1%2C2%7D%20%3D%20%5Cfrac%7B3%7D%7B4%7D%5Cpm%20%5Cfrac%7B%5Csqrt%7B329%7D%7D%7B4%7D)
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.