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(A) The given $x-t$ graph is a parabola. For $0 \le t \le T$,the graph is $x = kt^2$ (where $k > 0$),so velocity $v = dx/dt = 2kt$,which is a linear f

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(A) The given $x-t$ graph is a parabola. For $0 \le t \le T$,the graph is $x = kt^2$ (where $k > 0$),so velocity $v = dx/dt = 2kt$,which is a linear function of $t$ starting from $0$ and increasing. For $T \le t \le 2T$,the graph is a downward parabola,$x = -k'(t-2T)^2 + x_{max}$. The velocity $v = dx/dt = -2k'(t-2T) = 2k'(2T-t)$. This is a linear function that decreases to $0$ at $t = 2T$. Since speed is the magnitude of velocity $|v|$,the speed increases linearly from $0$ to a maximum value at $t = T$ and then decreases linearly to $0$ at $t = 2T$. This behavior is correctly represented by the graph in option $A$.

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

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