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(D) Given the differential equation: $\frac{dv}{dt} = 6 - 3v$.Rearranging the terms: $\frac{dv}{6 - 3v} = dt$.Integrating both sides: $\int \frac{dv}{

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All of the above.

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(D) Given the differential equation: $\frac{dv}{dt} = 6 - 3v$.Rearranging the terms: $\frac{dv}{6 - 3v} = dt$.Integrating both sides: $\int \frac{dv}{6 - 3v} = \int dt$.This yields: $-\frac{1}{3} \ln(6 - 3v) = t + C$.$\ln(6 - 3v) = -3t + C'$.At $t = 0$,$v = 0$,so $\ln(6) = C'$.Substituting $C'$: $\ln(6 - 3v) = -3t + \ln(6) \Rightarrow \ln(\frac{6 - 3v}{6}) = -3t$.Taking the exponential: $\frac{6 - 3v}{6} = e^{-3t} \Rightarrow 1 - 0.5v = e^{-3t} \Rightarrow v(t) = 2(1 - e^{-3t}) \, m/s$.Terminal speed $(t 	o \infty)$: $v = 2(1 - 0) = 2.0 \, m/s$.Initial acceleration $(t = 0)$: $a = \frac{dv}{dt} = 6 - 3(0) = 6.0 \, m/s^2$.Since all statements are correct,the answer is $(d)$.

The motion of a body is given by the equation $\frac{dv(t)}{dt} = 6.0 - 3v(t)$,where $v(t)$ is speed in $m/s$ and $t$ is in $s$. If the body was at rest at $t = 0$,which of the following statements is correct?

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