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: Set up the equation of motion.
The only force on the 1 kg object once gravity is ignored is the viscous drag \(D = \alpha v\), acting opposite to the motion. By Newton's second law,
\[ m\frac{dv}{dt} = -\alpha v \]
with \(m = 1\) kg and \(\alpha = 1\) kg/s.

Step 2: Separate variables and integrate.
Rearranging,
\[ \frac{dv}{v} = -\frac{\alpha}{m}\,dt \]
Integrating from the initial speed \(v_o\) at \(t=0\) to a general speed \(v\) at time \(t\),
\[ \ln\left(\frac{v}{v_o}\right) = -\frac{\alpha}{m}\,t \]

Step 3: Write the speed as a function of time.
\[ v(t) = v_o\, e^{-(\alpha/m)\,t} \]
This is a simple exponential decay of speed, with time constant \(\tau = m/\alpha = 1\) s.

Step 4: Apply the condition \(v = v_o/2\) and solve for \(t\).
\[ \frac{v_o}{2} = v_o\,e^{-t} \]
\[ \frac{1}{2} = e^{-t} \]
Taking the natural log of both sides,
\[ t = \ln 2 = 0.6931\ \text{s} \]

Final Answer:
Rounded to 2 decimal places, the object takes about 0.69 s to slow from \(v_o\) to \(v_o/2\).
\[ \boxed{t \approx 0.69\ \text{s}} \]