---
title: "Truck on circular road, speed and acceleration"
subject: "Applied Mechanics (Dynamics)"
chapter: 1
key: "semester_3/applied_mechanics_dynamics:truck-circular-road-tangential"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/truck-circular-road-tangential
site: BCT Engineering
---

# Truck on circular road, speed and acceleration

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

- Asked in: 2079 Bhadra, 4 marks
- Video: [12-125 | Curvilinear Motion | Engineering Dynamics Hibbeler 14th Edition | Engineers Academy](https://www.youtube.com/watch?v=h_AY7UbaAYQ) — Engineers Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/truck-circular-road-tangential
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-1

## Question

The truck travels along a circular road that has a radius of 50 m at a speed of 4 m/s. For a short distance when $t = 0$, its speed is then increased by $a_t = (0.4t)$ m/s$^2$, where $t$ is in seconds. Determine the speed and the magnitude of the truck's acceleration when $t = 4$ s.

## Short answer

Integrate a_t = 0.4t: v = 4 + 0.2t². At t=4 s: v = 7.2 m/s; a_t = 1.6 m/s²; a_n = v²/ρ = 7.2²/50 = 1.04 m/s²; a = √(1.6²+1.04²) = 1.91 m/s².

## Answer

On a circular path the acceleration has a tangential part $a_t=\dot v$ and a normal part $a_n=v^2/\rho$.

#### Speed at $t=4$ s
$$
v=v_0+\int_0^{4}0.4t\,dt=4+0.2t^{2}=4+0.2(4)^{2}=7.2\ \text{m/s}
$$

#### Acceleration at $t=4$ s
$$
\begin{aligned}
a_t&=0.4(4)=1.6\ \text{m/s}^2\\
a_n&=\frac{v^{2}}{\rho}=\frac{7.2^{2}}{50}=1.0368\ \text{m/s}^2\\
a&=\sqrt{a_t^{2}+a_n^{2}}=\sqrt{1.6^{2}+1.0368^{2}}=1.907\ \text{m/s}^2
\end{aligned}
$$

**Answer: $v=7.2$ m/s; $a=1.91$ m/s$^2$.**

## Check yourself

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

1. How is the speed at t = 4 s found?
   A. v = 4 + 0.4(4)
   B. v = 4 + integral of 0.4t dt = 4 + 0.2t^2
   C. v = 4 + 0.4t^2
   D. v = 4 x 0.4 x 4
   Answer: B. Integrating a_t = dv/dt from 0 to 4 s gives v = 4 + 0.2(16) = 7.2 m/s. (Key point: Speed from tangential acceleration)

2. How is the normal acceleration calculated?
   A. a_n = v^2 / rho using the speed at t = 4 s
   B. a_n = v^2 / rho using the initial 4 m/s
   C. a_n = 0.4t
   D. a_n = rho v^2
   Answer: A. a_n = 7.2^2/50 = 1.04 m/s^2 uses the instantaneous speed. (Key point: Normal acceleration)

3. What is a_t at t = 4 s?
   A. 0.4 m/s^2
   B. 1.6 m/s^2
   C. 7.2 m/s^2
   D. 0.2 m/s^2
   Answer: B. a_t = 0.4t = 0.4(4) = 1.6 m/s^2. (Key point: Tangential acceleration value)

4. How are a_t and a_n combined into the magnitude?
   A. a = a_t + a_n
   B. a = sqrt(a_t^2 + a_n^2)
   C. a = a_t - a_n
   D. a = a_t a_n
   Answer: B. They are perpendicular components, so sqrt(1.6^2 + 1.037^2) = 1.91 m/s^2. (Key point: Total acceleration)

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