## mathslover 3 years ago 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 ?

1. 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

2. mathslover

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

3. 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?

4. 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..

5. mathslover

yeah @DLS has a point\

6. ash2326

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

7. mathslover

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

8. ash2326

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

9. mathslover

|dw:1361020857966:dw|

10. 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 $\frac{dv}{d \theta}=0$ $\frac{d8(2 \sin \theta+\cos \theta)^-1}{d \theta}=0$

11. DLS

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

12. DLS

|dw:1361020933104:dw|

13. DLS

if my solution is close enough

14. mathslover

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

15. DLS

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

16. DLS

final thing

17. ash2326

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

18. DLS

@ash2326 I equated the times.. Tbus=Tboy

19. DLS

V=s/t

20. ash2326

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

21. ash2326

|dw:1361021378137:dw|

22. DLS

yeah i got ur query,hold on

23. 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

24. ash2326

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

25. mathslover

26. DLS

sure

27. ash2326

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

28. DLS

i thought that was 2 cosec theta :O

29. ash2326

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

30. DLS

:D

31. ash2326

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

32. ash2326

@mathslover do you have answer for this?

33. DLS

offline~

34. DLS

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

35. mathslover

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

36. anonymous

bahhhh!!!!..

37. anonymous

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

38. mathslover

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

39. anonymous

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

40. 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 =

41. ash2326

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

42. mathslover

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

43. DLS

hold on

44. mathslover

|dw:1361184323999:dw| ok :)

45. DLS

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

46. mathslover

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

47. mathslover

??

48. DLS

wait I got confused :P sec

49. DLS

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

50. DLS

$\LARGE \cot \theta=\frac{1}{\tan \theta}$

51. DLS

idk why i got confused !

52. DLS

$\LARGE \cot \theta=\frac{AC}{2}$

53. mathslover

OH got it thanks a lot

54. DLS

:P