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 , show that the pedestrian can cross the road safely in front of the car if

`u gt v sin (alpha)` , where `alpha = tan^-1 (w/l).`

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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=`Wsec(alpha)`

and B'C=`Wtan(alpha)`

Thus

`(Wsec(alpha))/u<=(I+Wtan(alpha))/v`

```vWsec(alpha)<=(u(Icos(alpha)+Wsin(alpha)))/cos(alpha)`

`vW<=uIcos(alpha)+uWsin(alpha)`

`W(v-usin(alpha))<=uIcos(alpha)`

`if`

`u>vsin(alpha)`

`-usin(alpha)<-vsin^2(alpha)`

`v-usin(alpha)<v-vsin^2(alpha)=vcos^2(alpha)`

`Wvcos^2(alpha)<=vIsin(alpha)cos(alpha)`

`vcos(alpha)(Isin(alpha)-Wcos(alpha))>=0`

`vcos(alpha)>=0`

``Since `0<=alpha<=pi/2` ,therefore

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

`tan(alpha)>=W/I`

`alpha>=tan^(-1)(W/I)`

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