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 ?
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ash2326
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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
mathslover
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well if we take the man to reach more safely there is that is the first case
ash2326
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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?
DLS
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|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
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yeah @DLS has a point\
ash2326
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But we need to find the minimum V for that, he has to walk straight
mathslover
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the man will make an angle theta
first we have to find the componenets ( x and y ) and then solve for theta
ash2326
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|dw:1361020702285:dw|
We aren't given D, we can't find angle without it
mathslover
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|dw:1361020857966:dw|
DLS
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\[v=\frac{d}{t} \]
\[BC=2cosec \theta\]
\[t_{bus}=t_{boy} \]
\[\frac{4+2\cot \theta}{8}=\frac{8}{v}=2\sin \theta+\cos \theta \]
\[v=\frac{8}{2\sin \theta+\cos \theta}\]
for v to be minimummmmmmmmmmm
\[\frac{dv}{d \theta}=0\]
\[\frac{d8(2 \sin \theta+\cos \theta)^-1}{d \theta}=0\]
DLS
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sorry i took the distance 4m correct that value..
DLS
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|dw:1361020933104:dw|
DLS
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if my solution is close enough
mathslover
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how r those angles : 2 cot theta and 2 cosec theta ?
DLS
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\[\LARGE \frac{12+2\cot \theta}{8}=\frac{8}{v}=\frac{6+\cot \theta}{4}\]
DLS
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final thing
ash2326
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How do you have 8/v?
man isn't travelling 8 meters
DLS
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@ash2326 I equated the times..
Tbus=Tboy
DLS
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V=s/t
ash2326
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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
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|dw:1361021378137:dw|
DLS
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yeah i got ur query,hold on
DLS
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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
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I solved this, there is no global minima for v.
We need to revisit the problem
mathslover
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got to go now,,,... sorry but please post your opinions
DLS
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sure
ash2326
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@DLS the man is travelling with an angle, his speed is v
I just found the hypotenuse distance
DLS
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i thought that was 2 cosec theta :O
ash2326
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Yes, if you solve the square root, you will get 2 cosec theta
DLS
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:D
ash2326
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I'm not convinced about the car travelling an extra distance 2 cot theta
ash2326
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@mathslover do you have answer for this?
DLS
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offline~
DLS
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Just do the whole thing for him :P From the beginning :D
mathslover
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yep --- 4/3 --- ans..
Mashy
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bahhhh!!!!..
Mashy
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doing all that other crap!! :P.. good thing you put up the answer!!
mathslover
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@Mashy you cn put up your soln... I posted it becz ash wnted to check the ans... Please post your soln
Mashy
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4/3 is the simple method.. of going straight :O..
DLBlast
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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
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@mathslover Use my first post, I had told you for minimum it has to be straight
mathslover
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DLS can you tell me how you got 2 cot theta there ?
DLS
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hold on
mathslover
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|dw:1361184323999:dw|
ok :)
DLS
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isnt the sol clear?
they are taking for it to be going straight
mathslover
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But I am trying to do for the angle theta... please tell how you got 2 cot theta there
mathslover
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??
DLS
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wait I got confused :P sec
DLS
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|dw:1361185375230:dw|
AC=2 cot theta
DLS
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\[\LARGE \cot \theta=\frac{1}{\tan \theta}\]
DLS
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idk why i got confused !
DLS
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\[\LARGE \cot \theta=\frac{AC}{2}\]
mathslover
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OH got it thanks a lot
DLS
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:P