The speeds of overtaking and overtaken vehicles are 80 and 50 kmph, respectively on a two-way traffic road. The average acceleration during overtaking may be assumed as $0.95 m / s e c^{2}$ . What is the minimum length (in m) of the overtaking zone?
[Assume reaction time for overtaking = 2.5 sec and length of vehicle = 6 m]

Open in App

Solution

Speed of overtaking vehicle,
$V = 80 k m p h = 22.22 m / s e c$

Speed of overtaken vehicle,
$V_{b} = 50 k m p h = 13.89 m / s e c$

Average acceleration during overtaking,
$a = 0.95 m / s e c^{2}$

Overtaking sight distance for two ways traffic,
$O S D = d_{1} + d_{2} + d_{3}$
$= (V_{b} t + V_{b} + 2 S + V T)$

$d_{1} = V_{b} t = 13.89 \times 2.5 = 34.7 m$

$d_{2} = V_{b} T + 2 S$

$S = 0.7 V_{b} + 6 = 0.7 \times 13.89 + 6 = 15.723 m$

$T = \sqrt{\frac{4 S}{a}} = \sqrt{\frac{4 \times 15.723}{0.95}} = 8.136 s e c$

$d_{2} = 13.89 \times 8.136 + 2 \times 15.723 = 144.5 m$

$d_{3} = V T = 22.22 \times 8.136 = 180.8 m$

$O S D = d_{1} + d_{2} + d_{3} = 34.7 + 144.5 + 180.8 = 360 m$

Now, minimum length overtaking zone
$= 3 \times O S D = 3 \times 360 = 1080 m$

Also, desirable length of overtaking zone
$= 5 \times O S D = 5 \times 360 = 1800 m$

Suggest Corrections

Similar questions

On one way traffic road, the speeds of overtaking and overtaken vehicles are 80 kmph and 50 kmph respectively. If the acceleration of the overtaking vehicle is 2.5 kmph per second, calculate the overtaking sight distance (Assume spacing between vehicles is 16 m, reaction time of driver is 2 seconds)

Q. The speed of overtaking vehicle is 90 kmph and acceleration is 3.0 kmph/sec. The desirable length of overtaking zone for one way traffic should be

[Assume the length of the vehicle as 6 m and reaction time as 2 sec]

The speed of overtaking and overtaken vehicles on a highway are $85 k m p h$ and $70 k m p h$ respectively. Calculate the overtaking sight-distance needed for two way traffic. Assume the acceleration of the overtaking vehicle as $2.5 k m p h$ per second and the speed of the vehicle in the opposite direction as $85 k m p h$

Q. For a two-way traffic road, the following are the particulars:
Speed of overtaking vehicle = 65 Kmph
Speed difference between the vehicles = 15 kmph
Acceleration of overtaking vehicle = 3.28 kmph/sec
Perception time of driver of overtaking vehicle = 2 seconds
Length of overtaking vehicle = 6 m
The length of safe overtaking sight distance will be

Q. For a highway with design speed of 100 kmph, the safe overtaking sight distance is (assume acceleration as

0.53 m / s^{2}

)

Join BYJU'S Learning Program