Applied Maths - Mechanics 19

Suppose the man cross the road at an angle A to the perpendicular path wrt to the direction of the road. The the optimum point is when the man just missed the front car as shown.
http://img204.imageshack.us/img204/4193/carddg.jpg
Let the perpendicular direction be y and the horizontal direction be x where motion directions are taken as positive.
Let the man be at a uniform speed of v, then
y = vcosAt
x = vsinAt
The time taken by the man to completely traverse the breath of the car is given by
b = vcosAt1 … (1)
By that amount of time the horizontal distance of the man has travelled is x = vsinAt1
In order for the man not to be knocked by the second car, for just miss
(a + vsinAt1) = ut1...(2)
(1)=> t1 = b/(vcosA)
Sub into (2),
a + vsinA(b)/vcosA = ub/vcosA
or avcosA + bvsinA = ub or v = ub/(acosA + bsinA) … (3)
where v and A are variables and a, b and u are constant.
Differentiate wrt A,
(dv/dA)acosA – avsinA + (dv/dA)bsinA + bvcosA = 0
dv/dA = v(asinA – bcosA)/(acosA + bsinA)
d2v/dA2 = v[(acosA + bsinA)(acosA + bsinA) – (asinA – bcosA)(-asinA + bcosA)] / (acosA + bsinA)2
d2v/dA2 = v(a2 + b2) / (acosA + bsinA)2 > 0
dv/dA = 0 => tanA = b/a which is the minimum for v
tanA = b/a, then cosA = a/√(a2 + b2) and sinA = b/√(a2+b2)
Put into (3), v = ub/(aa/√(a2 + b2) + bb/√(a2 + b2))
v = bu/√(a2 + b2)
Time to cross the road = [c/cosA]/v
= [c√(a2 + b2)/a][√(a2 + b2)/bu]
=c(a2 + b2)/abu
=c/u(a/b + b/a)