A-body-is-at-rest-at-x-0-At-t-0-it-starts-moving-in-the-positive-x-direction-with-a-constant-acceleration-At-the-same-instant-another-body-passes-through-x-0-moving-in-the-positive-x-directio

Question Number 16645 by Tinkutara last updated on 24/Jun/17

A body is at rest at x = 0 . At t = 0, it

starts moving in the positive x - direction

with a constant acceleration . At the

same instant another body passes

through x = 0 moving in the positive

x - direction with a constant speed . The

position of the first body is given by

x_{1} (t) after time^{'} t^{'} and that of the

second body by x_{2} (t) after the same

time interval . Which of the following

graphs correctly describes (x_{1} - x_{2}) as

a function of time^{'} t^{'} ?

Commented by Tinkutara last updated on 24/Jun/17

Answered by ajfour last updated on 24/Jun/17

only a little time after t = 0

x_{2} will be greater than x_{1} since

the first body will have gained

only a little velocity while the

second body had an initial

velocity of you, that is, if time

has not progressed by much

x_{1} - x_{2} < 0 . And after sufficient

time first body (because it is accelerated)

continues gaining velocity, surpases

velocity u and then overtakes the

second (x_{1} = x_{2}) and thereafter their separation

goes increasing [(x_{1} - x_{2}) increases] .

such is represented only in (2) .

Commented by Tinkutara last updated on 24/Jun/17

Thanks Sir!