Shoutbox

Ok here is a question that i heard. I have found the answers so could someone just reconfirm it cause it was in my friends question paper at his univ in his preliminary exams and he didnt know how to solve it so i showed it to him today
But there is no way to confirm the answers as the teacher is not available now ..... the holidays before the exams are on ....

Question:
A car is moving along a road's edge at a velocity of 8 m/s and is 2m wide.
If a man 12 m away starts to cross the road then what is the minimum velocity with which he should cross the road such that he wont be hit by the car. Also find the angle at which he crosses the road

Answers:
Velocity = 1.5291 m/s
Angle = 49.87 degrees (wrt to the road)

Tips
The man doest walk straight accross the road. He walks kinda inclined .... like this ....



I dont wanna tell my method but it involves basic calculus .... its easy but i dont wanna say anything as i dont wanna influence your method .... thats the only way we can get unbiased answers i guess

Just tell me what your answers are .... and what your basic method is ... no need to explain the steps lol Just say "we have A and B we do this and we get C" and so on

* John Anderton feels this is gonna take some time

I'll do it for ya. Took physics last year. just hold on tho

Basically, what I can see is that there is two equations:
(let's hope they're right!)
8t=sin(Θ; )Vt + 12
2=cos(Θ )Vt
solve for the minimization for V

hmm dodgy theta's

anyways, I solved for t in each equation, then set them equal to each other.

V = 8/(6cos() + sin() )

graphing this gives


and differentiating
dV/d = -8*(-6sin()+cos())/(6cos()+sin())^2

solving for critical points gives = arctan(1/6) or or .0029 degrees. The corresponding velocity comes out to be 1.315 m/s


ya, ya its different, but I might have messed up the equations at the beginning

quote:

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Originally posted by MeEtc
ya, ya its different, but I might have messed up the equations at the beginning

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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

oh err..this is too tough for me, i m just in 11th standard..but John Anderton, wat exactly was the exam?

i checked what you did and it seems to be ok

quote:

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Originally posted by Rubber Stamp
oh err..this is too tough for me, i m just in 11th standard..but John Anderton, wat exactly was the exam?

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Im doing engineering. You surely know what engineering in India is like And anyways ... how would you know calculus in 11th grade unless you are a JEE student .....
omgz l33tz0rz

quote:

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Originally posted by Chrono
i checked what you did and it seems to be ok

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You mean me rite ??