If a car appears in horison and within 2 seconds passes you by, whats the speed it's doing? While watching the first 4 seconds of driving at 745 km/h is ludicrous from any angle
wondered  


*

*If we knew the curvature of the Earth in a "flat" desert, what would be the speed of the car?

*Assuming we don't know the curvature of the Earth, how long would it take for the car to go around the Earth ( assuming a "flat" desert all the way travelling on a great arc)?
Using my ninja pause skills it seemed it took 2 seconds from the time the car appeared ( as a dot ) in horizon to the time it passed by the camera, although 2 second window of observation seems to lead to a great error in overall estimates, using the 745 as the value for 1 and 2 the error of human observation could be calculated (?)     
 A: In order to solve this question, we need to determine the distance from the eyes of the observer to the horizon, as this will be the relevant distance the car travels within those 2 seconds. Let's define a few variables:
$d =$ distance between observer and horizon
$v =$ velocity of the car
$t =$ time for car to travel from horizon to observer
$h =$ distance of observer's eyes above ground
$R =$ radius of Earth    
This calculation makes a few considerable approximations. First, it assumes that the Earth is spherical, and that the region between the observer and the horizon is a part of that sphere. Light is assumed to move in perfectly straight lines, neglecting any effects such as diffusion or refraction.
Here is a highly exaggerated diagram showing the Earth, and a 'not-to-scale' observer:

We can now create a right angle triangle, the three vertices located at the observer's eyes, the horizon (the end of the dotted line furthest from the observer), and the centre of the Earth. The sides of the triangle are $h+R$, $R$ and $d$. By a bit of Pythagoras' Theorem, it turns out that:
$$d = \sqrt{h^2+2hR}$$
The mean radius of the Earth, $R$, according to Wikipedia, is $(6.371 \times 10^6)m$, and let's assume that $h\approx 2m$. Therefore, $d\approx 5048m$.
Then, given that $t = 2s$, we can calculate the velocity of the car with:
$$v = \frac{d}{t} \approx 2524ms^{-1}$$
$$= 9086kmh^{-1}$$
So, with a velocity of around $9000kmh^{-1}$, and a Mach Number of around $7.4$, it goes without saying that we have stumbled upon a pretty absurd result. Which makes me question whether this deserves be an answer or a comment. Anyways, this value gives us a few possible scenarios to work with:


*

*The radius of curvature of the desert may be a lot smaller than that of the Earth, i.e. there might have been a hill

*The car can go ridiculously fast without becoming scrap metal (unlikely)

*The camera resolution/human observation cannot detect something crossing the literal horizon

*The camera is a mere few centimetres above the ground

*The effects of light diffusion and refraction are significant to the distance of the horizon

*The above theory and calculations are wrong

