A car starting from position of rest, moves with constant acceleration x. Then it moves with constant deceleration y and become stationary. If the total time elapsed during this is t, then the total distance travelled by car in time t is

Options

A. `((xy)/(x + y )) t `
B. `1/2 ((xy )/(x + y )) t ^(2)`
C. `2 ((xy)/(x + y )) t ^(2)`
D. `((xy)/(x + y )) t ^(2)`

Answer

B

Explanation

Supose a car strting from position of rest moves with constant acceleration x in time t gain maximum velocity v.
`therefore In v = v _(0) + at, v _(0) =0 , a = x and ` putting `t =t _(1)`
`therefore v = xt _(1)`
`therefore t _(1) = (v)/(x) " "...(1)`
After gaining maximum velocity, it moves with constant deceleration y in times `t_(2)` and become stationary then,
`therefore` Putting `v =v _(0) + at, v =0, v _(0) = v , a =- t and t = t _(2)`
`therefore 0=v -yt ^(2) ` in it
`therefore t _(2) = (v)/(y) " "...(2)`
Total time `t =t _(1) + t _(2)`
`therefore t = (v)/(x) + (v )/(y) = ((y+x)/( xy))v`
`therefore v = ((xy)/(x + y ))t `
Now average velocity `v.= (0+v)/(2) = 1/2 ((xy)/(x + y )) t`
`therefore ` Distance covered `d =v. xx t = 1/2 ((xy)/(x + y )) t ^(2)`