Question

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

• A. The terminal speed is 3.0 m/s
• B. The speed varies with the time as $v\left(t\right)=2(1-e^{-3t})m/s$
• C. The speed is 0.1 m/s when the acceleration is half the initial value
• D. The magnitude of the initial acceleration is 5.0 $m/s^2$

Answer 1

The correct option is B

The speed varies with the time as $v\left(t\right)=2(1-e^{-3t})m/s$

$\frac{dv}{dt}=6-3v\Rightarrow \frac{dv}{6-3v}=dt$
Integrating both sides,
$int\frac{dv}{6-3v}=\int dt\Rightarrow \frac{lo{g}_e(6-3v)}{-3}=t+K_1\Rightarrow lo{g}_e(6-3v)=-3t+K_2\dots \dots \left(i\right)$
$Att=0,v=0\therefore lo{g}_{e}6=K_2$
Substituting the value of $K_2$ in equation (i)
$lo{g}_e(6-3v)=-3t+lo{g}_{e}6\Rightarrow lo{g}_e\left(\frac{6-3v}{6}\right)=-3t\Rightarrow e^{-3t}=\frac{6-3v}{6}\Rightarrow 6-3v=6e^{-3t}\Rightarrow 3v=6(1-e^{-3t})\Rightarrow v=2(1-e^{-3t})\therefore v_{terminal}=2m/s(Whent=\infty )$
Acceleration,$a=\frac{dv}{dt}=\frac{d}{dt}\left[2\right(1-e^{-3t}\left)\right]=6e^{-3t}$
Initial acceleration $=6m/s^2$.