A body is at rest at x=0. At t=0,it starts mo | vedclass

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(C) For the body starting from rest at $x=0$ with constant acceleration $a$:$x_{1} = \frac{1}{2}at^{2}$For the body moving with constant speed $v$ sta

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(C) For the body starting from rest at $x=0$ with constant acceleration $a$:$x_{1} = \frac{1}{2}at^{2}$For the body moving with constant speed $v$ starting from $x=0$:$x_{2} = vt$Let $f(t) = x_{1} - x_{2} = \frac{1}{2}at^{2} - vt$.This is a quadratic equation in $t$ representing a parabola opening upwards.At $t=0$,$f(0) = 0$.The derivative is $f'(t) = at - v$.Setting $f'(t) = 0$ gives $t = \frac{v}{a}$.At $t = \frac{v}{a}$,the function reaches its minimum value: $f(\frac{v}{a}) = \frac{1}{2}a(\frac{v}{a})^{2} - v(\frac{v}{a}) = \frac{v^{2}}{2a} - \frac{v^{2}}{a} = -\frac{v^{2}}{2a}$.Since the minimum value is negative and the parabola opens upwards,the graph starts at the origin,goes below the $t$-axis,reaches a minimum at $t = \frac{v}{a}$,and then increases,crossing the $t$-axis at $t = \frac{2v}{a}$.This matches the graph shown in the solution image.

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