Question

$The \: relation \: between \: position \\ (x) \: and \: time \: (t) \: are \: given \: below \\ for \: a \: particle \: moving \: along \: \\ a \: straight \: line. \: \\ Which \: of \: the \: following \\ equation \: represents \: uniformly \: \\ accelerated \: motion? \: \\ (where \: \alpha \: and \: \beta \: are \: positive \: \\ constants) \\ \\ 1) \: \beta x =\alphat +\alpha \beta \\ 2) \: \alpha x =\beta+ t \\ 3) \: xt =\alpha \beta \\ 4) \: \alpha t = \sqrt{ \beta + x}$

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Answer 1

If a particle undergoes uniformly accelerated motion, its displacement as a function of time can be given by the second equation of motion.
$s=ut+\dfrac{1}{2}at^2$
Displacement is the change in position of the particle.
Therefore, position of particle as a function of time can be expressed as follows:
$\Rightarrow x-x_0=ut+\dfrac{1}{2}at^2\\ \Rightarrow x(t)=x_0+ut+\dfrac{1}{2}at^2$
As the motion of the particle is accelerated, $a\ne 0$.
Therefore $x(t)$ must be a two-degree polynomial in terms of $t$.
Check all options whether they are two degree polynomials on not.
$1)\ \beta x=\alpha \beta \\ \Rightarrow x=\alpha \qquad \text{Nope}\\ 2)\ \alpha x=\beta +t\\ \Rightarrow x=\dfrac{\beta}{\alpha}+\dfrac{1}{\alpha}t\qquad \text{Nope}\\ 3)\ xt=\alpha \beta \Rightarrow x=\alpha\beta t^{-1}\qquad \text{Nope}\\ 4)\ \alpha t=\sqrt{\beta +x} \Rightarrow x={\alpha}^2t^2-\beta\qquad \checkmark$

Answer 2

as you can see in only 4th option a is not equal to 0. and it is also not function of time . there for it is uniform acceralation. if you like this ans then please mark as brainlist