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


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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

	$Thanks Sir!$
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