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A body starts from rest and travels a distance S with uniform acceleration, then moves uniformly a distance 2S and finally comes to rest after moving further 5S under uniform retardation. Find the ratio of average velocity to maximum velocity : |
Answer» Solution :Graphically : Area of `(V-t)` curve represent displacement `S=(1)/(2)V_(max)t_(1)ort_(1)=(2S)/(V_(max))`, `2S=V_(max)t_(2)ort_(2)=(2S)/(V_(max))` `5S=(1)/(2)V_(max)t_(3)ort_(3)=(10S)/(V_(max))` `V_(av)=("Total displacement")/("Total time"),V_(av)=(S+2S+5S)/((2S)/(V_(max))+(2S)/(V_(max))+(10S)/(V_(max)))` `(V_(av))/(V_(max))=(8S)/(14S)=(4)/(7)(OR)` `(V_(av))/(V_(max))=overset("Total displacement")underset(2(("total displacement"),("during ACCELERATION"),("and retardation"))+(("Displacement"),("During uniform"),("velocity")))rarr` `therefore (V_(av))/(V_(max))=(8S)/(2(S+5S)+2S)=(8)/(14)=(4)/(7)` |
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