A 2m wide car is moving uniformly with a speed of 8m/s along the edge of a straight horizontal road. A pedestriun starts to cross a road with a speed "v" when the car is 12 m away from him. The minimum value of v for the pedestrian to cross the road safety is ?

- mathslover

- schrodinger

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- ash2326

There can be two cases, either the man crosses the road before the car reach him or he crosses it after the car has crossed

- mathslover

well if we take the man to reach more safely there is that is the first case

- ash2326

see the diagram|dw:1361020408653:dw|
Time taken for car to reach the man,
\[12 m\div 8 m/s= 1.5 seconds\]
In this time the man should cover the 2 meter distance, fairly simple now
Can you find v?

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## More answers

- DLS

|dw:1361020539994:dw|
@ash2326 just trying..i dont think there will be 2 cases,he should move making an angle theta with the axis not straight..

- mathslover

yeah @DLS has a point\

- ash2326

But we need to find the minimum V for that, he has to walk straight

- mathslover

the man will make an angle theta
first we have to find the componenets ( x and y ) and then solve for theta

- ash2326

|dw:1361020702285:dw|
We aren't given D, we can't find angle without it

- mathslover

|dw:1361020857966:dw|

- DLS

\[v=\frac{d}{t} \]
\[BC=2cosec \theta\]
\[t_{bus}=t_{boy} \]
\[\frac{4+2\cot \theta}{8}=\frac{8}{v}=2\sin \theta+\cos \theta \]
\[v=\frac{8}{2\sin \theta+\cos \theta}\]
for v to be minimummmmmmmmmmm
\[\frac{dv}{d \theta}=0\]
\[\frac{d8(2 \sin \theta+\cos \theta)^-1}{d \theta}=0\]

- DLS

sorry i took the distance 4m correct that value..

- DLS

|dw:1361020933104:dw|

- DLS

if my solution is close enough

- mathslover

how r those angles : 2 cot theta and 2 cosec theta ?

- DLS

\[\LARGE \frac{12+2\cot \theta}{8}=\frac{8}{v}=\frac{6+\cot \theta}{4}\]

- DLS

final thing

- ash2326

How do you have 8/v?
man isn't travelling 8 meters

- DLS

@ash2326 I equated the times..
Tbus=Tboy

- DLS

V=s/t

- ash2326

I know that but man has to travel \[\sqrt{2^2+(2\cot \theta)^2}\]
so it should be
\[\frac{\sqrt{2^2+(2\cot \theta)^2}}{v}\]

- ash2326

|dw:1361021378137:dw|

- DLS

yeah i got ur query,hold on

- DLS

i took that 12+2cot theta
that is the total distance he has to travel i guess distance is scalar :o so why are u adding it vectorally? im getting confused too :S

- ash2326

I solved this, there is no global minima for v.
We need to revisit the problem

- mathslover

got to go now,,,... sorry but please post your opinions

- DLS

sure

- ash2326

@DLS the man is travelling with an angle, his speed is v
I just found the hypotenuse distance

- DLS

i thought that was 2 cosec theta :O

- ash2326

Yes, if you solve the square root, you will get 2 cosec theta

- DLS

:D

- ash2326

I'm not convinced about the car travelling an extra distance 2 cot theta

- ash2326

@mathslover do you have answer for this?

- DLS

offline~

- DLS

Just do the whole thing for him :P From the beginning :D

- mathslover

yep --- 4/3 --- ans..

- anonymous

bahhhh!!!!..

- anonymous

doing all that other crap!! :P.. good thing you put up the answer!!

- mathslover

@Mashy you cn put up your soln... I posted it becz ash wnted to check the ans... Please post your soln

- anonymous

4/3 is the simple method.. of going straight :O..

- anonymous

before i can start i need to know the in physics what is the start point for all formula is = to ? or 0.00 or what the name of all start point plz? start point =

- ash2326

@mathslover Use my first post, I had told you for minimum it has to be straight

- mathslover

DLS can you tell me how you got 2 cot theta there ?

- DLS

hold on

- mathslover

|dw:1361184323999:dw|
ok :)

- DLS

isnt the sol clear?
they are taking for it to be going straight

- mathslover

But I am trying to do for the angle theta... please tell how you got 2 cot theta there

- mathslover

??

- DLS

wait I got confused :P sec

- DLS

|dw:1361185375230:dw|
AC=2 cot theta

- DLS

\[\LARGE \cot \theta=\frac{1}{\tan \theta}\]

- DLS

idk why i got confused !

- DLS

\[\LARGE \cot \theta=\frac{AC}{2}\]

- mathslover

OH got it thanks a lot

- DLS

:P

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