---
title: "Tangential and normal components of velocity"
subject: "Applied Mechanics"
chapter: 8
key: "semester_1/applied_mechanics:tangential-normal-components-velocity"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/tangential-normal-components-velocity
site: BCT Engineering
---

# Tangential and normal components of velocity

Applied Mechanics · Chapter 8: Kinematics of Particles and Rigid Body

- Asked in: 2071 Chaitra, 4 marks
- Video: [Curvilinear Motion: Normal and Tangential components (Learn to solve any problem)](https://www.youtube.com/watch?v=hwHy4Dewc7g) — Question Solutions
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/tangential-normal-components-velocity
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-8

## Question

What are tangential and normal components of velocity? Explain with examples.

## Short answer

Velocity is always tangent to the path, so its tangential component equals the speed v = ds/dt and its normal component is zero. For a car at constant 20 m/s on a 100 m radius track, v_t = 20 m/s, v_n = 0.

## Answer

#### Tangential and normal components
When a particle moves along a curved path, its velocity vector is **always tangent to the path**. Choose a moving frame at the particle with unit vectors e_t (along the tangent, in the direction of motion) and e_n (perpendicular to it, toward the centre of curvature).

- **Tangential component of velocity:** v_t = v = ds/dt, the speed of the particle along the path.
- **Normal component of velocity:** v_n = 0, because the velocity has no component perpendicular to the path.

So v = v e_t.

For the acceleration the normal and tangential components are both present:
- a_t = dv/dt (rate of change of speed)
- a_n = v²/ρ (rate of change of direction, ρ = radius of curvature)

#### Examples
1. **Car on a circular track of radius 100 m moving at constant 20 m/s:** v_t = 20 m/s, v_n = 0, a_t = 0, a_n = 20²/100 = 4 m/s² toward the centre.
2. **Car on the same track speeding up at 2 m/s² when its speed is 20 m/s:** a_t = 2 m/s², a_n = 4 m/s², total a = √(2² + 4²) = 4.47 m/s².
3. **A projectile at its highest point:** velocity is horizontal (tangent), v_t = v_x; the whole acceleration g is normal, so ρ = v_x²/g.

## Check yourself

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

1. What is the normal component of velocity?
   A. v^2/ρ
   B. Zero, since velocity is always tangent to the path
   C. Equal to the tangential component
   D. dv/dt
   Answer: B. v = v e_t, so v_n = 0. (Key point: Normal component of velocity)

2. Which gives the acceleration components?
   A. a_t = v^2/ρ, a_n = dv/dt
   B. a_t = dv/dt, a_n = v^2/ρ
   C. a_t = v/ρ, a_n = dv/dt
   D. Both equal dv/dt
   Answer: B. Tangential acceleration is the rate of change of speed, normal is v^2/ρ. (Key point: Acceleration components)

3. A car speeds up at 2 m/s^2 at 20 m/s on a track of 100 m radius. What is total a?
   A. 2 m/s^2
   B. 6 m/s^2
   C. 4 m/s^2
   D. 4.47 m/s^2
   Answer: D. a_n = 400/100 = 4 and a_t = 2, so a = sqrt(4 + 16) = 4.47 m/s^2. (Key point: Car on track)

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- [Normal, tangential acceleration; radius of curvature](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aposition-vector-normal-tangential-radius-curvature) — Applied Mechanics, chapter 8

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