mathslover
  • mathslover
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
schrodinger
  • schrodinger
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ash2326
  • 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
  • mathslover
well if we take the man to reach more safely there is that is the first case
ash2326
  • 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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DLS
  • 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
  • mathslover
yeah @DLS has a point\
ash2326
  • ash2326
But we need to find the minimum V for that, he has to walk straight
mathslover
  • mathslover
the man will make an angle theta first we have to find the componenets ( x and y ) and then solve for theta
ash2326
  • ash2326
|dw:1361020702285:dw| We aren't given D, we can't find angle without it
mathslover
  • mathslover
|dw:1361020857966:dw|
DLS
  • 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
  • DLS
sorry i took the distance 4m correct that value..
DLS
  • DLS
|dw:1361020933104:dw|
DLS
  • DLS
if my solution is close enough
mathslover
  • mathslover
how r those angles : 2 cot theta and 2 cosec theta ?
DLS
  • DLS
\[\LARGE \frac{12+2\cot \theta}{8}=\frac{8}{v}=\frac{6+\cot \theta}{4}\]
DLS
  • DLS
final thing
ash2326
  • ash2326
How do you have 8/v? man isn't travelling 8 meters
DLS
  • DLS
@ash2326 I equated the times.. Tbus=Tboy
DLS
  • DLS
V=s/t
ash2326
  • 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
  • ash2326
|dw:1361021378137:dw|
DLS
  • DLS
yeah i got ur query,hold on
DLS
  • 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
  • ash2326
I solved this, there is no global minima for v. We need to revisit the problem
mathslover
  • mathslover
got to go now,,,... sorry but please post your opinions
DLS
  • DLS
sure
ash2326
  • ash2326
@DLS the man is travelling with an angle, his speed is v I just found the hypotenuse distance
DLS
  • DLS
i thought that was 2 cosec theta :O
ash2326
  • ash2326
Yes, if you solve the square root, you will get 2 cosec theta
DLS
  • DLS
:D
ash2326
  • ash2326
I'm not convinced about the car travelling an extra distance 2 cot theta
ash2326
  • ash2326
@mathslover do you have answer for this?
DLS
  • DLS
offline~
DLS
  • DLS
Just do the whole thing for him :P From the beginning :D
mathslover
  • mathslover
yep --- 4/3 --- ans..
anonymous
  • anonymous
bahhhh!!!!..
anonymous
  • anonymous
doing all that other crap!! :P.. good thing you put up the answer!!
mathslover
  • mathslover
@Mashy you cn put up your soln... I posted it becz ash wnted to check the ans... Please post your soln
anonymous
  • anonymous
4/3 is the simple method.. of going straight :O..
anonymous
  • 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
  • ash2326
@mathslover Use my first post, I had told you for minimum it has to be straight
mathslover
  • mathslover
DLS can you tell me how you got 2 cot theta there ?
DLS
  • DLS
hold on
mathslover
  • mathslover
|dw:1361184323999:dw| ok :)
DLS
  • DLS
isnt the sol clear? they are taking for it to be going straight
mathslover
  • mathslover
But I am trying to do for the angle theta... please tell how you got 2 cot theta there
mathslover
  • mathslover
??
DLS
  • DLS
wait I got confused :P sec
DLS
  • DLS
|dw:1361185375230:dw| AC=2 cot theta
DLS
  • DLS
\[\LARGE \cot \theta=\frac{1}{\tan \theta}\]
DLS
  • DLS
idk why i got confused !
DLS
  • DLS
\[\LARGE \cot \theta=\frac{AC}{2}\]
mathslover
  • mathslover
OH got it thanks a lot
DLS
  • DLS
:P

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