Question

The motion of a body is given by the equation$\frac{dv}{dt}=6-3v$ where $v$ is the speed in $m{s}^{-1}$ and $t$ istime in $s$ . The body is at rest at $t=0.$ The speedvaries with time as

• A. $v=(1-e^{-3t})$
• B. $v=2(1-e^{-3t})$
• C. $v=(1+e^{-2t})$
• D. $v=2(1+e^{-2t})$

Answer 1

The correct option is B $v=2(1-e^{-3t})$

$\begin{array}{c}Given\frac{dv}{dt}=6-3v\\ ={\int }_0^u\frac{dv}{6-3v}={\int }_0^{t}dt\\ t=-\frac{1}{3}{\left[v\left(6-3v\right)\right]}_0^{\theta }\\ -3t=\ln \left(6-3v\right)-\ln 6\\ -3t=\ln \left(\frac{6-3v}{6}\right)=\ln \left(1-\frac{v}{2}\right)\\ 1-\frac{v}{2}=e^{-3t}\\ 2-v=2e^{-3t}\\ v=2\left(1-e^{-3t}\right)\end{array}$

Option B.