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(B) Given $\frac{dK}{dt} = \gamma t$. Since $K = \frac{1}{2}mv^2$,we have $\frac{dK}{dt} = mv \frac{dv}{dt} = \gamma t$.Thus,$v \frac{dv}{dt} = \frac{

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$(A), (B), (D)$

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(B) Given $\frac{dK}{dt} = \gamma t$. Since $K = \frac{1}{2}mv^2$,we have $\frac{dK}{dt} = mv \frac{dv}{dt} = \gamma t$.Thus,$v \frac{dv}{dt} = \frac{\gamma t}{m}$.Integrating both sides,$\int v dv = \int \frac{\gamma}{m} t dt$,which gives $\frac{v^2}{2} = \frac{\gamma t^2}{2m}$.Therefore,$v = \sqrt{\frac{\gamma}{m}} t$. This shows that speed is proportional to time,so $(B)$ is true.The acceleration $a = \frac{dv}{dt} = \sqrt{\frac{\gamma}{m}}$,which is a constant. Since $F = ma$,the force $F = \sqrt{\gamma m}$ is also constant,so $(A)$ is true.Since $v = \frac{dx}{dt} = \sqrt{\frac{\gamma}{m}} t$,integrating gives $x = \int \sqrt{\frac{\gamma}{m}} t dt = \frac{1}{2} \sqrt{\frac{\gamma}{m}} t^2$. The distance increases quadratically with time,not linearly,so $(C)$ is false.Since the force is constant,it is a conservative force,so $(D)$ is true.Thus,statements $(A), (B),$ and $(D)$ are true.

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