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(D) Given the speed $v = \alpha \sqrt{x}$.Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = \alpha \sqrt{x}$,which implies $\int x^{-1/2} dx = \int \

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(D) Given the speed $v = \alpha \sqrt{x}$.Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = \alpha \sqrt{x}$,which implies $\int x^{-1/2} dx = \int \alpha dt$.Integrating both sides,we get $2\sqrt{x} = \alpha t$,so $x = \frac{\alpha^2 t^2}{4}$.The tangential acceleration is $a_t = \frac{dv}{dt} = \frac{d}{dt}(\alpha \sqrt{x}) = \alpha \cdot \frac{1}{2\sqrt{x}} \cdot \frac{dx}{dt} = \alpha \cdot \frac{1}{2\sqrt{x}} \cdot (\alpha \sqrt{x}) = \frac{\alpha^2}{2}$.Since $\alpha$ is a constant,$a_t$ is constant.The centripetal acceleration is $a_c = \frac{v^2}{r} = \frac{(\alpha \sqrt{x})^2}{r} = \frac{\alpha^2 x}{r}$.Substituting $x = \frac{\alpha^2 t^2}{4}$,we get $a_c = \frac{\alpha^2}{r} \cdot \frac{\alpha^2 t^2}{4} = \frac{\alpha^4}{4r} t^2$.Thus,$a_t$ is constant and $a_c$ is proportional to $t^2$ (a parabolic curve starting from the origin).Comparing this with the given options,the graph where $a_t$ is a horizontal line and $a_c$ is a parabola starting from the origin is the correct one.

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