Answer

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**Hint:**In order to solve this question, we will compare force formulas in terms of acceleration and given form to find relation between acceleration and time and then using integration we will find velocity relation with time and determine the nature of the curve between these variables.

**Formula used:**

The acceleration is written in derivative form as $a = \dfrac{{dv}}{{dt}}$ and $v = \dfrac{{dx}}{{dt}}$ where v is velocity, dx is change in displacement of the body and ‘dt’ is change in time.

**Complete answer:**

According to the question, we have given that force is increasing with time with the relation $F = \alpha t$ and as we know from newton’s second law of motion that force is given by $F = ma$ where m is mass and a is acceleration of the body so, comparing both the values of force we get,

$ma = \alpha t$ since, m and $\alpha $ are both positive constants so, we see that acceleration is directly proportional to the time as $a \propto t$ so, Curve $1$ which is a straight line represents acceleration against time.

So, Options (A) is the correct answer

Now, as we have, $a \propto t$ we can write for some constant it as $a = kt$ and we know the derivative form of acceleration in terms of velocity ‘dv’ and time ‘dt’ as $a = \dfrac{{dv}}{{dt}}$ so we get,

$\dfrac{{dv}}{{dt}} = kt$

Now, integrating both sides we get,

$\int {dv} = k\int {tdt} $

using integration formula $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $

$ \Rightarrow v = \dfrac{{k{t^2}}}{2}$

so, velocity is directly proportional to square of time as $v \propto {t^2}$ whose graph will be a curved path like a parabola as shown in the given graph in curve $2$, so Curve $2$ represent the velocity against time.

So, Option (B) is the correct answer.

**Hence, the correct options are Option (A) and (B).**

**Note:**While solving such questions, always remember the nature of basic curves like straight lines, parabola and elliptical curves and also remember the integration formulas and derivative forms of acceleration and velocity to solve such questions efficiently.

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