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 $\frac{m}{s}$ and t in sec. If body was at rest at t = 0

• A. The terminal speed is $3.0\frac{m}{s}$
• B. The speed varies with the time as $v\left(t\right)=2(1-e^{-3t})\frac{m}{s}$
• C. The speed is $0.1\frac{m}{s}$ when the acceleration is half the initial value
• D. The magnitude of the initial acceleration is $0.5\frac{m}{s^2}$

Answer 1

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

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