The position-time (x-t) graph for positive ac | vedclass

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Question

$x=\frac{1}{2} a t^2 \quad$ (if coefficients of $t^2$ is positive $(a > 0)$ upward opening parabola)

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Vedclass · Answer

$x=\frac{1}{2} a t^2 \quad$ (if coefficients of $t^2$ is positive $(a > 0)$ upward opening parabola)

Graph	Characteristics
$(A)$	$(i)$ has $v > 0$ and $a < 0$ throughout
$(B)$	$(ii)$ has $x > 0,$ throughout and has a point with $v = 0$ and a point with $a = 0$
$(C)$	$(iii)$ has a point with zero displacement for $t > 0$
$(D)$	$(iv)$ has $v < 0$ and $a > 0$

The position-time $(x-t)$ graph for positive acceleration is