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(C) Given the equation of motion: $x = \frac{1}{2} vt$. Since velocity $v = \frac{dx}{dt}$, we substitute this into the equation: $x = \frac{1}{2} \le

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acceleration $f$ is constant

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(C) Given the equation of motion: $x = \frac{1}{2} vt$. Since velocity $v = \frac{dx}{dt}$, we substitute this into the equation: $x = \frac{1}{2} \left( \frac{dx}{dt} \right) t$. Rearranging the terms to separate variables: $\frac{2 dt}{t} = \frac{dx}{x}$. Integrating both sides: $2 \int \frac{dt}{t} = \int \frac{dx}{x} \implies 2 \ln |t| + C' = \ln |x|$. This simplifies to $\ln |t^2| + C' = \ln |x|$, which implies $x = c t^2$ for some constant $c$. Now, find the velocity $v$ by differentiating $x$ with respect to $t$: $v = \frac{dx}{dt} = \frac{d}{dt} (c t^2) = 2ct$. Finally, find the acceleration $f$ by differentiating $v$ with respect to $t$: $f = \frac{dv}{dt} = \frac{d}{dt} (2ct) = 2c$. Since $2c$ is a constant, the acceleration $f$ is constant.

The law of motion of a body moving along a straight line is $x = \frac{1}{2} vt$, where $x$ is its distance from a fixed point on the line at time $t$ and $v$ is its velocity. Then:

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