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mathslover
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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 ?
 one year ago
 one year ago
mathslover Group Title
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 ?
 one year ago
 one year ago

This Question is Closed

ash2326 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
well if we take the man to reach more safely there is that is the first case
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.0
see the diagramdw: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?
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
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..
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
yeah @DLS has a point\
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.0
But we need to find the minimum V for that, he has to walk straight
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
the man will make an angle theta first we have to find the componenets ( x and y ) and then solve for theta
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.0
dw:1361020702285:dw We aren't given D, we can't find angle without it
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
dw:1361020857966:dw
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
\[v=\frac{d}{t} \] \[BC=2cosec \theta\] \[t_{bus}=t_{boy} \] \[\frac{4+2\cot \theta}{8}=\frac{8}{v}=2\sin \theta+\cos \theta \] \[v=\frac{8}{2\sin \theta+\cos \theta}\] for v to be minimummmmmmmmmmm \[\frac{dv}{d \theta}=0\] \[\frac{d8(2 \sin \theta+\cos \theta)^1}{d \theta}=0\]
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
sorry i took the distance 4m correct that value..
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
dw:1361020933104:dw
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
if my solution is close enough
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
how r those angles : 2 cot theta and 2 cosec theta ?
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
\[\LARGE \frac{12+2\cot \theta}{8}=\frac{8}{v}=\frac{6+\cot \theta}{4}\]
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.0
How do you have 8/v? man isn't travelling 8 meters
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
@ash2326 I equated the times.. Tbus=Tboy
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.0
I know that but man has to travel \[\sqrt{2^2+(2\cot \theta)^2}\] so it should be \[\frac{\sqrt{2^2+(2\cot \theta)^2}}{v}\]
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.0
dw:1361021378137:dw
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
yeah i got ur query,hold on
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.0
I solved this, there is no global minima for v. We need to revisit the problem
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
got to go now,,,... sorry but please post your opinions
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.0
@DLS the man is travelling with an angle, his speed is v I just found the hypotenuse distance
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
i thought that was 2 cosec theta :O
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.0
Yes, if you solve the square root, you will get 2 cosec theta
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.0
I'm not convinced about the car travelling an extra distance 2 cot theta
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.0
@mathslover do you have answer for this?
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
Just do the whole thing for him :P From the beginning :D
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
yep  4/3  ans..
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
bahhhh!!!!..
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
doing all that other crap!! :P.. good thing you put up the answer!!
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
@Mashy you cn put up your soln... I posted it becz ash wnted to check the ans... Please post your soln
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
4/3 is the simple method.. of going straight :O..
 one year ago

DLBlast Group TitleBest ResponseYou've already chosen the best response.0
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 =
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.0
@mathslover Use my first post, I had told you for minimum it has to be straight
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
DLS can you tell me how you got 2 cot theta there ?
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
dw:1361184323999:dw ok :)
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
isnt the sol clear? they are taking for it to be going straight
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
But I am trying to do for the angle theta... please tell how you got 2 cot theta there
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
wait I got confused :P sec
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
dw:1361185375230:dw AC=2 cot theta
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
\[\LARGE \cot \theta=\frac{1}{\tan \theta}\]
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
idk why i got confused !
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
\[\LARGE \cot \theta=\frac{AC}{2}\]
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
OH got it thanks a lot
 one year ago
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