question_answer

A car traveling at a certain speed\[{{v}_{1}}\]overtakes another car travelling at a slower speed\[{{v}_{2}}\]at A. An hour later the faster car reverses its direction, changes it's speed to\[{{v}_{3}}\]and after some time meets the slower car now coming in the opposite direction at position B. The distance between the positions A and B is 10 km. Find the speed of the slower car if\[{{V}_{1}}-{{V}_{2}}={{V}_{3}}+{{V}_{2}}\]

Answer:

  Let\[{{V}_{1}}>{{V}_{2}}\]and let Car \[1({{V}_{3}})\] meet Car \[2({{V}_{2}})\] after a time t. From figure, \[AC=AB+BC\] \[\Rightarrow {{V}_{1}}\times 1={{V}_{2}}(1+t)+{{V}_{3}}t\Rightarrow t=1hr\] \[\therefore 10={{V}_{2}}(1+t)\Rightarrow {{V}_{2}}=5km/hr.\]