---
title: "Path of particle in velocity field"
subject: "Fluid Mechanics"
chapter: 4
key: "semester_3/fluid_mechanics:particle-path-velocity-field"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/particle-path-velocity-field
site: BCT Engineering
---

# Path of particle in velocity field

Fluid Mechanics · Chapter 4: Hydrokinematics

- Asked in: 2079 Baisakh, 6 marks
- Video: [Streamline and Trajectory Solving](https://www.youtube.com/watch?v=Tqp2Ygt6Sp8) — Matt B
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/particle-path-velocity-field
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-4

## Question

Given the velocity field $\vec V = (5x)\hat i + (15y+11)\hat j + (19t^2)\hat k$ m/s. Determine the path of particle which is at (4,6,2) m at time t = 3 s.

## Short answer

Parametric path: x=4e^{5(t-3)}, y=6.733e^{15(t-3)}−0.733, z=2+6.333(t³−27). Eliminating t yields y=0.1052x³−0.733.

## Answer

The path of a particle is found by integrating $dx/dt = u$, $dy/dt = v$, $dz/dt = w$ with the initial position at the given time, then eliminating $t$.

**Given:** $u = 5x$, $v = 15y + 11$, $w = 19t^2$; at $t = 3$ s the particle is at $(4, 6, 2)$.

**x-motion**

$$
\frac{dx}{x} = 5\,dt \;\Rightarrow\; x = 4\,e^{5(t-3)}
$$

**y-motion**

$$
\begin{aligned}
\frac{dy}{15y + 11} &= dt\\
\ln(15y + 11) &= 15t + C\\
y + \tfrac{11}{15} &= \left(6 + \tfrac{11}{15}\right)e^{15(t-3)} = 6.733\,e^{15(t-3)}
\end{aligned}
$$

**z-motion**

$$
z = 2 + \frac{19}{3}\left(t^3 - 27\right)
$$

**Parametric path** (parameter $t$):

$$
x = 4e^{5(t-3)},\quad y = 6.733e^{15(t-3)} - 0.733,\quad z = 2 + 6.333(t^3 - 27)
$$

**Eliminating t:** from $x$, $e^{5(t-3)} = x/4$ and $t = 3 + \tfrac{1}{5}\ln(x/4)$. Hence

$$
y = 6.733\left(\frac{x}{4}\right)^{3} - 0.733 = 0.1052\,x^3 - 0.733
$$

$$
z = 2 + 6.333\left[\left(3 + \tfrac{1}{5}\ln\tfrac{x}{4}\right)^3 - 27\right]
$$

**Answer: the path is the curve $y = 0.1052x^3 - 0.733$ in the xy-plane, with $z = 2 + 6.333\,(t^3 - 27)$ and $t = 3 + 0.2\ln(x/4)$.** It passes through $(4, 6, 2)$; the particle moves away from the origin as $t$ increases.

## Check yourself

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

1. How is the path of a particle found from a velocity field?
   A. Integrate dx/dt=u, dy/dt=v, dz/dt=w with the known position at a given time, then eliminate t
   B. Solve dy/dx=v/u at one fixed instant
   C. Differentiate the velocity to get acceleration and set it to zero
   D. Set u=v=w=0 and solve for x, y, z
   Answer: A. A path follows one particle in time, so the velocity components are integrated with the initial position, then t is eliminated. (Key point: Method for a path)

2. For u = 5x with x=4 at t=3 s, what is x(t)?
   A. x = 4 + 5(t-3)
   B. x = 4 e^(5(t-3))
   C. x = 4 e^(5t)
   D. x = 20 e^(t-3)
   Answer: B. dx/x = 5 dt gives ln x = 5t + C; using x=4 at t=3 gives x = 4e^(5(t-3)). (Key point: x-motion)

3. Why does the y-solution have the form y + 11/15 = 6.733 e^(15(t-3))?
   A. Because v is constant so y grows linearly
   B. Because v depends on t^2, giving a cubic
   C. dy/(15y+11)=dt integrates to a log, and the constant is fixed using y=6 at t=3
   D. Because y must equal x at all times
   Answer: C. Integrating gives ln(15y+11)=15t+C; at y=6, 15y+11=101, so (y+11/15) = (6+11/15)e^(15(t-3)). (Key point: y-motion)

4. Which relation between y and x results after eliminating t?
   A. y = 0.1052 x^5 - 0.733
   B. y = 6.733 x - 0.733
   C. y = 0.1052 x^(1/3) - 0.733
   D. y = 0.1052 x^3 - 0.733
   Answer: D. Since e^(5(t-3)) = x/4, e^(15(t-3)) = (x/4)^3, so y = 6.733(x/4)^3 - 0.733. (Key point: Eliminating t)

## Related in Fluid Mechanics

- [Streamline and time, particle x = 1 to 2](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Astreamline-time-ax-ayx2-x1-2) — Fluid Mechanics, chapter 4
- [Pathline and streamlines for velocity field](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Apathline-streamlines-bx-t0) — Fluid Mechanics, chapter 4
- [Stagnation points in 2D velocity field](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Astagnation-points-velocity-field) — Fluid Mechanics, chapter 4
- [Streamline and time in $A/x$, $Ay/x^2$ field](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Astreamline-time-ax-ayx2-x1-3) — Fluid Mechanics, chapter 4
- [Streamline, $\psi$ and $\phi$ for given field](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Astreamline-psi-phi-2x3-6x2y) — Fluid Mechanics, chapter 4
- [Terminal velocity of steel sphere in glycerine](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Asphere-terminal-velocity-glycerine) — Fluid Mechanics, chapter 9

## Related in other subjects

- [Vibrating particle velocity, acceleration and path](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Avibrating-particle-x-100-sin-pi-t) — Applied Mechanics, chapter 8
- [Dependent motion; radius of curvature of particle path](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Adependent-motion-particle-ax-ay) — Applied Mechanics, chapter 8
- [Velocity and acceleration of a particle](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Avelocity-and-acceleration-of-a-particle) — Mathematics III, chapter 2
- [Velocity and acceleration of a particle](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Avelocity-and-acceleration-of-a-particle-two) — Mathematics III, chapter 2