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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 describes $(x_{1}-x_{2})$ as a function of time $t$ ?
Option: 1

Option: 2

Option: 3

Option: 4

Answers (1)

best_answer

For 1st body:-

As u = 0 and a=constant

v₁= 0+at⇒v₁=at

$x_1=\frac{1}{2}at^2$

For 2nd body:-

since v is constant, so a=dv/dt=0

$x_2=vt$

So, $x_1-x_2=\frac{1}{2}at^2-vt----------------(1)$

Now, to find the nature of graph, we will differentiate the equation with respect to t:-

$\frac{d(x_1-x_2)}{dt}=at-v$

If we equate this to zero, then t=v/a

In equation 1, if t=0 and t=2v/a, then $x_1-x_2=0$

and if t<2v/a, then slope $\frac{d(x_1-x_2)}{dt}$ wiil be negative

if t=v/a, then $\frac{d(x_1-x_2)}{dt}$ will be zero

if t>=2v/a, slope $\frac{d(x_1-x_2)}{dt}$ will be positive

Therefore it is a parabola after crossing x-axis again. Curve (c) satisfies this .

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Nehul

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