question_answer

The driver of a train moving at a speed\[{{\upsilon }_{1}}\], sights another train at a distance d, ahead of him moving in the same direction with a slower speed\[{{\upsilon }_{2}}\]. He applies the brakes and gives a constant retardation a to his train. Show that there will be no collision if\[\text{d}>{{\left( {{\upsilon }_{\text{1}}}-{{\upsilon }_{\text{2}}} \right)}^{\text{2}}}/\text{2a}\].

Answer:

                For no collision of two trains, the relative velocity of faster train w.r.t slower train \[\left( {{\upsilon }_{\text{1}}}-{{\upsilon }_{\text{2}}} \right)\] should become zero, while travelling a relative displacement d with acceleration a. Here ; \[\text{u}={{\upsilon }_{\text{1}}}-{{\upsilon }_{\text{2}}};\text{ }\upsilon =0;\text{a}=-\text{a };\text{S}=\text{d}\]. As. \[{{\upsilon }^{\text{2}}}={{\text{u}}^{\text{2}}}+\text{2as}\], so, \[0={{\left( {{\upsilon }_{\text{1}}}-\text{ }{{\upsilon }_{\text{2}}} \right)}^{\text{2}}}+\text{ 2}(-\text{a})\text{d}\]or \[\text{d}={{({{\upsilon }_{\text{1}}}-{{\upsilon }_{\text{2}}})}^{\text{2}}}/\text{2a}\] There will be no collision if\[\text{d}>{{({{\upsilon }_{\text{1}}}-{{\upsilon }_{\text{2}}})}^{\text{2}}}/\text{2a}\].