A body starts from rest and travels a distance x with uniform acceleration, then it travels a distance 2x with uniform speed, finally it travels a distance 3x with uniform retardation and comes to rest. If the complete motion of the particle is along a straight line, then the ratio of its average velocity to maximum velocity is

Options

A. `(6)/(7)`
B. `(4)/(5)`
C. `(3)/(5)`
D. `(2)/(5)`

Answer

C

Explanation

Area of (v-t) graph=displacement
For uniform acceleration `s=(1)/(2)xxt_(1)xxv_("max")`
`t_(1)=(2s)/(v_("max"))`....(1)
For uniform speed `2s=t_(2)xxv_("max")`
`t_(2)=(2s)/(v_("max")` ......(2)
For uniform retardation `3s=(1)/(2)xxt_(3)xxv_("max")`
`t_(3)=(6s)/(v_("max"))` .....(3)
`|vecv_(avg)|=("Displacement")/("Total time")=(s+2s+3s)/(t_(1)+t_(2)+t_(3))`
`(6s)/((2s)/(v_("max"))+(2s)/v_("max")+(6s)/(v_("max")))=(6s)/((10s)/(v_("max")))=(6v_("max"))/(10)`
`|vecv_(avg)|=(6)/(10)xxv_("max")` then `(|vecv_("max")|)/(v_("max"))=(3)/(5)`