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

A car starting from position of rest, moves with constant acceleration

1.	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
Answer» `((xy)/(x + y )) t ` `1/2 ((xy )/(x + y )) t ^(2)` `2 ((xy)/(x + y )) t ^(2)` `((xy)/(x + y )) t ^(2)` Solution :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)`