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Finding the best angle for maximum chance to cross the road in time.

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In Georgian (ქართული): https://drive.google.com/file/d/0B_4P3RNorJzTN3h0Z3cwaG04SXc

Real life math problem and solution

Crossing The Road As A Pedestrian:

The car has n times more speed than the pedestrian. What is the best α angle from the perpendicular to maximize the chance of crossing the road in time before the car approaches?

VSCode Blockman Icon

Restrictions:

1) n > 1

2) 0 ≤ α < 90°

3) The width of the road is unknown

4) The distance between the car and the pedestrian is unknown

Solution:

In order to get the best angle with maximum probability of crossing the road before the car approaches ---- when the pedestrian crosses the road, the distance between the car and him/her must be maximum:

First, define the variables:

Speed of the pedestrian — V

Speed of the car — n⋅V

Horizontal speed of the pedestrian — Vx

Vertical speed of the pedestrian — Vy

Horizontal distance between the first and the last point of the pedestrian — a (not α)

Horizontal distance between the car and the pedestrian — b

Width of the road — d

Time spent to cross the road — t

math formula

math formula

math formula

b affects the maximum, but it does not affect the best angle to get maximum, so we can remove it.

math formula

math formula

d affects the maximum, but it does not affect the best angle to get maximum, so we can remove it.

math formula

math formula

In order to find the extremum of a function, we must equalize its derivative to zero:

math formula

cos2(α) cannot be zero, so α cannot be 90°, and therefore:

math formula

And finally, α = arcsin(1/n) because we already knew that 0 ≤ α < 90° and n > 1.

Answer:

α = arcsin(1/n)

n (n > 1) α (0 ≤ α < 90°)
1.1 ≈65.380023°
1.5 ≈41.810315°
2 =30°
3 ≈19.471221°
4 ≈14.477512°
5 ≈11.536959°
10 ≈5.73917°
15 ≈3.822554°
20 ≈2.865984°
50 ≈1.145992°
100 ≈0.572967°
1000 ≈0.057296°
Calculator used


Geogebra plot:

y = b + d ⋅ tan(α) - n⋅V⋅d/(V ⋅ cos(α))

and

α = arcsin(1/n)

https://www.geogebra.org/m/YV7g2YKA

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