Question

An object of mass 1 kg is launched with an initial speed of \(v_o\) into a large tank filled with a viscous liquid. The liquid exerts a resistive force (drag) of the form \(D = \alpha v\) on any object that is moving inside it, where \(v\) is the instantaneous speed of the object and \(\alpha = 1\) kg/s. If the effect of gravity is ignored, the time taken by the object to slow down to the speed \(v_o/2\) is ________ s (rounded off to 2 decimal places). Assume that the tank is sufficiently large for the above deceleration to happen inside the tank.

Hint:

Set up m dv/dt = -alpha v and separate variables to get exponential decay; find t when v = v0/2.

Answer 1

Correct Answer: 0.69

Step 1: Recognize the form of the equation.
With gravity ignored, the drag $D=\alpha v$ is the only force, so Newton's second law gives $m\,dv/dt=-\alpha v$. This has exactly the same mathematical form as a discharging RC circuit or a first-order cooling process, so it can be treated as a standard exponential-decay problem with time constant $\tau = m/\alpha$.

Step 2: Find the time constant.
$\tau = m/\alpha = 1\text{ kg} / 1\text{ kg/s} = 1$ s.
For this type of decay, the general solution is $v(t) = v_o e^{-t/\tau}$.

Step 3: Use the half-life relation.
For any exponential decay $v(t)=v_o e^{-t/\tau}$, the time to fall to a fraction $f$ of the starting value satisfies $f = e^{-t/\tau}$, so $t = -\tau \ln f$. Here the object must slow to half its speed, so $f = 1/2$:
\[ t = -\tau \ln(1/2) = \tau \ln 2 \]

Step 4: Substitute numbers.
$t = 1 \times \ln 2 = 1 \times 0.6931 = 0.6931$ s.
Rounded to 2 decimal places, $t \approx 0.69$ s.
\[ \boxed{t \approx 0.69\ \text{s}} \]