---
title: "Bullet in viscous medium, velocity and position"
subject: "Applied Mechanics (Dynamics)"
chapter: 1
key: "semester_3/applied_mechanics_dynamics:bullet-viscous-medium"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/bullet-viscous-medium
site: BCT Engineering
---

# Bullet in viscous medium, velocity and position

Applied Mechanics (Dynamics) · Chapter 1: Curvilinear Motion of Particles

- Asked in: 2078 Kartik, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/bullet-viscous-medium
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-1

## Question

A bullet is fired into a viscous medium with an initial velocity of 80 m/s. The resistance of the medium produces a resistance equal of $a = -0.5v^3$ m/s$^2$, where $v$ is in m/s. Calculate the bullet's velocity and position 3 sec after it is fired.

## Short answer

Integrate a = dv/dt = -0.5v³ to get v(t). At t = 3 s, velocity is 0.577 m/s, and integrating again yields position s = 3.44 m.

## Answer

The acceleration depends on velocity, so use $a=\dfrac{dv}{dt}$ and then $v=\dfrac{ds}{dt}$.

#### Velocity after 3 s
$$
\begin{aligned}
\frac{dv}{dt}&=-0.5v^{3}\\
\int_{80}^{v}\frac{dv}{v^{3}}&=-0.5\int_0^{t}dt\\
\left[-\frac{1}{2v^{2}}\right]_{80}^{v}&=-0.5t\\
\frac{1}{v^{2}}&=\frac{1}{80^{2}}+t
\end{aligned}
$$

At $t=3$ s:

$$
\frac1{v^2}=\frac1{6400}+3=3.000156\ \Rightarrow\ v=0.5773\ \text{m/s}
$$

#### Position after 3 s
$$
v=\frac{ds}{dt}=\left(\frac1{6400}+t\right)^{-1/2}
$$

$$
s=\int_0^{3}\left(\frac1{6400}+t\right)^{-1/2}dt=2\left[\sqrt{\tfrac1{6400}+t}\right]_0^{3}=2(1.7321-0.0125)=3.439\ \text{m}
$$

**Answer: $v\approx0.577$ m/s and $s\approx3.44$ m at $t=3$ s.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Because a depends on v, which form of acceleration is used first?
   A. a = dv/dt and separate variables
   B. a = v dv/ds only
   C. Constant acceleration formulas
   D. a = d^2s/dt^2 with s constant
   Answer: A. Constant-acceleration equations do not apply; separate variables in dv/dt = -0.5v^3. (Key point: Choice of method)

2. What is the result of integrating dv/v^3 = -0.5 dt from 80 m/s?
   A. 1/v^2 = 1/6400 + t
   B. 1/v = 1/80 + t
   C. v = 80 - 0.5t
   D. v^2 = 6400 - t
   Answer: A. The integral of v^-3 is -1/(2v^2), giving 1/v^2 = 1/80^2 + t. (Key point: Integrated result)

3. What is the velocity at t = 3 s?
   A. 80 m/s
   B. 0.577 m/s
   C. 3.44 m/s
   D. 1.73 m/s
   Answer: B. 1/v^2 = 3.000156, so v = 0.577 m/s. (Key point: Velocity at 3 s)

4. How is position found after the velocity function is known?
   A. Differentiate v again
   B. Integrate v = ds/dt from 0 to 3 s
   C. Use s = v t
   D. Use s = 80 t
   Answer: B. s = integral of (1/6400 + t)^(-1/2) dt = 2[sqrt(1/6400 + t)] from 0 to 3 = 3.44 m. (Key point: Position)

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