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 ?

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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 ?

Physics
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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
well if we take the man to reach more safely there is that is the first case
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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Other 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..
yeah @DLS has a point\
But we need to find the minimum V for that, he has to walk straight
the man will make an angle theta first we have to find the componenets ( x and y ) and then solve for theta
|dw:1361020702285:dw| We aren't given D, we can't find angle without it
|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
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
How do you have 8/v? man isn't travelling 8 meters
  • DLS
@ash2326 I equated the times.. Tbus=Tboy
  • DLS
V=s/t
I know that but man has to travel \[\sqrt{2^2+(2\cot \theta)^2}\] so it should be \[\frac{\sqrt{2^2+(2\cot \theta)^2}}{v}\]
|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
I solved this, there is no global minima for v. We need to revisit the problem
got to go now,,,... sorry but please post your opinions
  • DLS
sure
@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
Yes, if you solve the square root, you will get 2 cosec theta
  • DLS
:D
I'm not convinced about the car travelling an extra distance 2 cot theta
@mathslover do you have answer for this?
  • DLS
offline~
  • DLS
Just do the whole thing for him :P From the beginning :D
yep --- 4/3 --- ans..
bahhhh!!!!..
doing all that other crap!! :P.. good thing you put up the answer!!
@Mashy you cn put up your soln... I posted it becz ash wnted to check the ans... Please post your soln
4/3 is the simple method.. of going straight :O..
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 =
@mathslover Use my first post, I had told you for minimum it has to be straight
DLS can you tell me how you got 2 cot theta there ?
  • DLS
hold on
|dw:1361184323999:dw| ok :)
  • DLS
isnt the sol clear? they are taking for it to be going straight
But I am trying to do for the angle theta... please tell how you got 2 cot theta there
??
  • 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
\[\LARGE \cot \theta=\frac{AC}{2}\]
OH got it thanks a lot
  • DLS
:P

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