Question 87: Motion in 1D

Question

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 $${x_1}\left( t \right)$$ after time $$'t';$$ and that of the second body by $${x_2}\left( t \right)$$ after the same time interval. Which of the following graphs correctly describes $$\left( {{x_1} - {x_2}} \right)$$ as a function of time $$'t'$$ ?

Accepted Answer

For the body starting from rest
$${x_1} = 0 + {1 \over 2}a{t^2} \Rightarrow {x_1} = {1 \over 2}a{t^2}$$
For the body moving with constant speed
$${x_2} = vt$$
$$	herefore$$ $${x_1} - {x_2} = {1 \over 2}a{t^2} - vt$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {{d\left( {{x_1} - {x_2}} ight)} \over {dt}} = at - v$$
at $$t=0,$$ $$\,\,\,\,\,\,{x_1} - {x_2} = 0$$, so graph should start from origin.
For $$at  v;$$ the slope is positive. So $${x_1} - {x_2}$$ > 0 it means here velocity of first body is greater than second body.
We know the relation between distance and time is.
$$S = ut + {1 \over 2}a{t^2}$$, which is a equation parabola. So the graph should be a parabola.
These characteristics are represented by graph $$(b).$$

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Motion in 1D

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