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(C) Given acceleration $a = \frac{dv}{dt} = bt$.Integrating with respect to time $t$:$\int dv = \int bt \, dt \Rightarrow v = \frac{1}{2}bt^2 + C_1$.A

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${v_0}t + \frac{1}{6}b{t^3}$

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(C) Given acceleration $a = \frac{dv}{dt} = bt$.Integrating with respect to time $t$:$\int dv = \int bt \, dt \Rightarrow v = \frac{1}{2}bt^2 + C_1$.At $t = 0$,$v = v_0$,so $C_1 = v_0$.Thus,the velocity is $v = \frac{1}{2}bt^2 + v_0$.Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = \frac{1}{2}bt^2 + v_0$.Integrating with respect to time $t$:$x = \int (\frac{1}{2}bt^2 + v_0) dt = \frac{1}{2}b(\frac{t^3}{3}) + v_0t + C_2$.At $t = 0$,$x = 0$,so $C_2 = 0$.Therefore,the distance travelled is $x = v_0t + \frac{1}{6}bt^3$.

The acceleration of a particle is increasing linearly with time $t$ as $bt$. The particle starts from the origin with an initial velocity ${v_0}$. The distance travelled by the particle in time $t$ will be

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