---
title: "Overtaking car acceleration numerical"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:overtaking-acceleration-numerical-2-01"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/overtaking-acceleration-numerical-2-01
site: BCT Engineering
---

# Overtaking car acceleration numerical

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2082 Baisakh, 8 marks
- Video: [2nd Numerical of Overtaking Sight Distance - Highway Geometric Design - Transportation Engineering 1](https://www.youtube.com/watch?v=8c0wC1BgSA4) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/overtaking-acceleration-numerical-2-01
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

An impatient car driver stuck behind a truck travelling at 35 kmph decides to overtake the truck. The acceleration characteristics of car is given by, dV/dt = 2.01 − 0.05V where V is speed in m/s and time is in sec.

i) What is acceleration after 20 and 40 sec?
ii) What is maximum speed attainable by car?
iii) How far will car travel in 260 sec?

## Short answer

Solve dV/dt = 2.01 - 0.05V from 9.722 m/s. Acceleration is 0.561 m/s2 at 20 s and 0.206 m/s2 at 40 s; maximum speed 40.2 m/s (144.7 km/h); distance in 260 s is 9842 m.

## Answer

The acceleration falls linearly with speed, so $\dfrac{dV}{dt} = 2.01 - 0.05V$ is solved by separating variables. Assumption: the car starts at the truck's speed, $V_0 = 35$ km/h $= 9.722$ m/s at $t = 0$.

#### Speed as a function of time
$$
\int \frac{dV}{2.01-0.05V} = \int dt \;\Rightarrow\; -20\ln(2.01-0.05V) = t + C
$$

At $t=0$, $V=V_0$. This gives

$$
V(t) = 40.2 - (40.2 - 9.722)\,e^{-0.05t} = 40.2 - 30.478\,e^{-0.05t}\ \text{m/s}
$$

and the acceleration is $a = 2.01 - 0.05V = 0.05(40.2 - V_0)e^{-0.05t} = 1.5239\,e^{-0.05t}$.

#### (i) Acceleration after 20 s and 40 s
$$
\begin{aligned} V(20) &= 40.2 - 30.478e^{-1} = 28.988\ \text{m/s}, & a(20) &= 2.01 - 0.05(28.988) = 0.561\ \text{m/s}^2\\ V(40) &= 40.2 - 30.478e^{-2} = 36.075\ \text{m/s}, & a(40) &= 2.01 - 0.05(36.075) = 0.206\ \text{m/s}^2 \end{aligned}
$$

#### (ii) Maximum speed attainable
Speed is maximum when $dV/dt = 0$:

$$
V_{max} = \frac{2.01}{0.05} = 40.2\ \text{m/s} = 144.7\ \text{km/h}
$$

(This is approached asymptotically; a real car would be limited by the speed limit and road.)

#### (iii) Distance travelled in 260 s
$$
\begin{aligned} x &= \int_0^{260} V\,dt = 40.2t - \frac{30.478}{0.05}\left(1 - e^{-0.05t}\right)\\ &= 40.2(260) - 609.56\left(1 - e^{-13}\right) = 9842.4\ \text{m} \end{aligned}
$$

**Answer:** (i) $a_{20} = 0.561$ m/s$^2$, $a_{40} = 0.206$ m/s$^2$; (ii) $V_{max} = 40.2$ m/s (144.7 km/h); (iii) distance $\approx$ **9842 m** (9.84 km).

## Check yourself

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

1. How is dV/dt = 2.01 - 0.05V solved for speed against time?
   A. Separate variables and integrate, using V0 = 9.722 m/s at t = 0
   B. Assume constant acceleration of 2.01
   C. Differentiate once
   D. Set V = 0
   Answer: A. Separation of variables gives V(t) = 40.2 - 30.478 e^(-0.05t) with the truck's speed as initial speed. (Key point: Method of solving dV/dt)

2. How is the maximum attainable speed found?
   A. Set t = 0
   B. Set dV/dt = 0, so V = 2.01/0.05 = 40.2 m/s
   C. Set V = 35 km/h
   D. Set acceleration to 2.01
   Answer: B. Maximum speed occurs when acceleration is zero, giving 40.2 m/s (144.7 km/h), approached asymptotically. (Key point: Maximum speed)

3. After 20 s the car speed is 28.988 m/s. What is the acceleration?
   A. 2.01 m/s^2
   B. 0.561 m/s^2
   C. 1.45 m/s^2
   D. 0 m/s^2
   Answer: B. a = 2.01 - 0.05(28.988) = 0.561 m/s^2. (Key point: Acceleration at a time)

4. How is distance travelled in 260 s obtained?
   A. Multiply 35 km/h by 260 s
   B. Integrate V dt from 0 to 260, giving about 9842 m
   C. Divide V_max by 260
   D. Use v^2 = 2as with a = 2.01
   Answer: B. Distance is the integral of speed over time: 40.2(260) - 609.56(1 - e^-13), about 9842 m. (Key point: Distance from speed)

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