The relation between position (x) and time (t | vedclass

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Question

(D) For uniformly accelerated motion,the position $(x)$ as a function of time $(t)$ must be a quadratic equation of the form $x = ut + \frac{1}{2}at^2

Vedclass · Accepted Answer

$\alpha t = \sqrt{\beta + x}$

Vedclass · Answer

(D) For uniformly accelerated motion,the position $(x)$ as a function of time $(t)$ must be a quadratic equation of the form $x = ut + \frac{1}{2}at^2$,where $u$ is the initial velocity and $a$ is the constant acceleration.Let us analyze option $(d)$:$\alpha t = \sqrt{\beta + x}$Squaring both sides:$(\alpha t)^2 = \beta + x$$\alpha^2 t^2 = \beta + x$$x = \alpha^2 t^2 - \beta$Comparing this with the standard kinematic equation $x = ut + \frac{1}{2}at^2$,we see that the position $x$ is a quadratic function of time $t$ (with $u = 0$ and $a = 2\alpha^2$).Since the acceleration $a = 2\alpha^2$ is a constant,this equation represents uniformly accelerated motion.Therefore,the correct option is $(d)$.

The relation between position $(x)$ and time $(t)$ is given below for a particle moving along a straight line. Which of the following equations represents uniformly accelerated motion? [where $\alpha$ and $\beta$ are positive constants]

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