A motor car of width **w** moves uniformly along a straight road , parallel to the pavement almost touching it. A pedestrian on the edge of the pavement at a distance **l** ahead of the car begins to walk uniformly to cross the road . If v is the speed of the car and u is the speed of the pedestrian relative to the road.

If`u = v sin (alpha)` , show that the pedestrian can cross the road just in front of the car , by walking relative to the road in a direction making an angle `(pi)/ 2 - (alpha)` with the direction of the motion of the car relative to the road.

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Let A be position of the car and B is position of the man.The distance between A and B is I (given).Let B' is point on the opposite side of the point B.Let man crosses road by making an angle `alpha` with BB' and reaches point C .Thus

BC=`W sec(alpha)`

and B'C=`W tan(alpha)`

Thus

`(W sec(alpha))/u=(I+W tan(alpha))/v`

`W/(vsin(alpha)cos(alpha))=(I cos(alpha)+Wsin(alpha))/(vcos(alpha))`

`W/sin(alpha)=Icos(alpha)+Wsin(alpha)`

`Wcos^2(alpha)=Icos(alpha)sin(alpha)`

`cos(alpha)(Wcos(alpha)-Isin(alpha))=0`

`either`

`cos(alpha)=0`

`alpha=pi/2`

`` It will not possible otherwise he will never cross the road.

`or`

`Wcos(alpha)=Isin(alpha)`

`tan(alpha)=W/I`

Thus pedestrian must move by making an angle (pi/2-alpha) with the direction of motion of car.

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