Question

A coronavirus droplet of mass 1 microgram ejects from the mouth of a patient with a velocity of 0.7 m/s and travels through air. The gravitational force experienced by it can be neglected due to the buoyancy effect. However, the droplet experiences air drag force proportional to its velocity and the drag coefficient is given as 1.0 μN-s/m. The distance travelled by the droplet before its velocity drops to 10% of its initial velocity (in m, rounded off to two decimal places) is \(\underline{\hspace{2cm}}\).

Hint:

For objects experiencing air drag, use the exponential decay of velocity and integrate to find the distance traveled.

Answer 1

Correct Answer: 0

The drag force on the droplet is given by:
\[ F_{\text{drag}} = c v \] Where: - \( c = 1.0 \, \mu\text{N·s/m} \) (drag coefficient), - \( v \) is the velocity of the droplet. The velocity decreases according to the following equation:
\[ v(t) = v_0 e^{-kt} \] Where:
- \( v_0 = 0.7 \, \text{m/s} \) (initial velocity),
- \( v(t) \) is the velocity at time \( t \),
- \( k \) is a constant related to the drag coefficient.
The velocity reduces to 10% of its initial velocity, i.e., \( v(t) = 0.1 v_0 \). Using this, we can solve for the time \( t \) it takes for this change: \[ 0.1 v_0 = v_0 e^{-kt} $\Rightarrow$ e^{-kt} = 0.1 $\Rightarrow$ t = \frac{\ln(10)}{k} \] Now, to calculate the distance traveled \( d \), we integrate the velocity:
\[ d = \int_0^t v_0 e^{-kt} dt = \frac{v_0}{k} \] Substituting the known values, we find that the distance traveled is approximately \( 0.01 \, \text{m} \).
Thus, the distance is approximately \( 0.01 \, \text{m} \).