The position x of a particle varies with time | vedclass

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(B) The given equation is $x = ct^2 + b$. This is the equation of a parabola.At $t = 0$,$x = b$. Since $b$ is a positive constant,the graph must inter

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(B) The given equation is $x = ct^2 + b$. This is the equation of a parabola.At $t = 0$,$x = b$. Since $b$ is a positive constant,the graph must intersect the $x$-axis at a positive value $b$ (i.e.,the $y$-intercept is positive).As $t$ increases,$x$ increases quadratically with $t$. The slope of the $x-t$ graph is given by $v = \frac{dx}{dt} = 2ct$. Since $c > 0$,the velocity $v$ is positive and increases linearly with time,meaning the graph is concave upwards.Therefore,the graph is a parabola that opens upwards and has a positive $y$-intercept at $x = b$.

The position $x$ of a particle varies with time $t$ as $x = ct^2 + b$,where $c$ and $b$ are positive constants. Which of the following graphs is correct?

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