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2.Motion in Straight Line
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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 describe $\left(x_{1}-x_{2}\right)$ as a function of time $t$?
A
B
C
D
(AIEEE-2008)
Solution
$\begin{array}{l}\,For\,the\,body\,starting\,from\,rest\\
{x_1} = 0 + \frac{1}{2}a{t^2} \Rightarrow {x_1} = \frac{1}{2}a{t^2}\\
For\,the\,body\,moving\,with\,{\rm{constant}}\,speed\\
{x_2} = vt\\
\therefore \,{x_1} – {x_2} = \frac{1}{2}a{t^2} – \,vt \Rightarrow \frac{{d\left( {{x_1} – {x_2}} \right)}}{{dt}} = at – v\\
at\,\,t = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x_1} – {x_2} = 0\\
For\,t < \frac{v}{a};\,the\,slope\,is\,negative\,\\
For\,t = \frac{v}{a};\,the\,slope\,is\,zero\,\\
For\,t > \frac{v}{a};\,the\,slope\,is\,positive
\end{array}$
Standard 11
Physics
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