The x-t graph of a particle moving along a st | vedclass

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(B) The $x-t$ graph is a parabola opening downwards,which can be represented by the equation $x = at - bt^2$,where $a$ and $b$ are positive constants.

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(B) The $x-t$ graph is a parabola opening downwards,which can be represented by the equation $x = at - bt^2$,where $a$ and $b$ are positive constants.The velocity $v$ is the derivative of position $x$ with respect to time $t$:$v = \frac{dx}{dt} = \frac{d}{dt}(at - bt^2) = a - 2bt$.This equation $v = a - 2bt$ represents a straight line with a negative slope $(-2b)$ and a positive intercept $(a)$.At $t = 0$,$v = a$ (positive).At $t = T$,the slope of the $x-t$ graph is zero,so $v = 0$. This gives $a - 2bT = 0$,or $a = 2bT$.At $t = 2T$,$v = a - 2b(2T) = a - 4bT = 2bT - 4bT = -2bT = -a$ (negative).Thus,the velocity decreases linearly from a positive value to zero at $t = T$ and then becomes negative,reaching $-a$ at $t = 2T$. This corresponds to the graph shown in option $B$.

The $x-t$ graph of a particle moving along a straight line is shown in the figure. The $v-t$ graph of the particle is correctly shown by:

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