A body is at rest at x=0. At t=0, it starts moving in the positiv… - Vidyadip

MCQ

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
✓
D

✓

Answer

Correct option: C.

c
$\begin{array}{l}
\,For\,the\,body\,starting\,from\,rest\\
{x_1} = 0 + \frac{1}{2}a{t^2} \Rightarrow {x_1} = \frac{1}{2}a{t^2}\\
For\,the\,body\,moving\,with\,{\rm{constant}}\,speed\\
{x_2} = vt\\
\therefore \,{x_1} - {x_2} = \frac{1}{2}a{t^2} - \,vt \Rightarrow \frac{{d\left( {{x_1} - {x_2}} \right)}}{{dt}} = at - v\\
at\,\,t = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x_1} - {x_2} = 0\\
For\,t < \frac{v}{a};\,the\,slope\,is\,negative\,\\
For\,t = \frac{v}{a};\,the\,slope\,is\,zero\,\\
For\,t > \frac{v}{a};\,the\,slope\,is\,positive
\end{array}$

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