quote:
Originally posted by MeEtc
ya, ya its different, but I might have messed up the equations at the beginning
Yes you did
What i did was that the man crosses the road at an angle alpha. So let the horizontal distance that he covers be x while the vertical distance that he covers (which is given) is 2m.
So the car covers distance 12+x with velocity 8m/s while the man covers distance (x^2+4)^1/2 (pythagoras theorem if anyone is still wondering) with velocity v
The time remains constant. So we have
(12+x)/8 = [ (x^2+4)^1/2 ]/v
v = 8 [(x^2+4)^1/2] / (12+x)
Differentiating wrt to x and since v should be minimum, its 1st derrivative is zero (cause for any function if its value is max or min at any point then the slope at that point is zero; thats the only basic calculus / math you need to know)
On solving we get the value of x.
Time t = (12+x)/8
We get time t.
v = [(x^2+4)^1/2] / t
We get velocity v
And finally angle (alpha) ......
2 = v sin(alpha) * t
alpha = sin^-1 [ 2/(vt) ]
Thats what i did
Usually the sums are much much tougher but when you are under pressure in the exam while solving those sums, you may not get the solution to such easy problems and thats what happened to my friend
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