A partical starts from rest and moves in a straight line. It travels a distance L with uniform acceleration and then moves with constant velocity a further distance `2L`. Finally, it comes to rest after moving a further distance 4L under uniform retardation. Find the ratio of average speed to the maximum speed.

Explanation

`A` to `B`: (constant acceleration)
`L=((0+v_(m))/2) t_(1)impliest_(1)=(2L)/v_(m)`
`B` to `C`: (constant velocity)
`2L=v_(m)t_(2)impliest_(2)=(2L)/v_(m)`
`C` to `D`: (constant retardation)
`4L=((v_(m)+0)/2)t_(3)impliest_(3)=(8L)/v_(m)`
Total distance= `L+2L+4L=7L`
Total time `=t_(1)+t_(2)_t_(3)=(2L)/v_(m)+(2L)/v_(m)+(8L)/v_(m)=(12L)/v_(m)`
Average speed `bar(v)=(7L)/((12L)/v_(m))=(7v_(m))/12`
`bar(v)/v_(m)=7/12`