Civil Engineering - Highway Engineering - Discussion

9. 

Over taking time required for a vehicle with design speed 50 km ph and overtaking acceleration 1.25 m/sec2 to overtake a vehicle moving at a speed 30 km ph, is

[A]. 5.0 secs
[B]. 6.12 secs
[C]. 225.48 secs
[D]. 30 secs

Answer: Option B

Explanation:

No answer description available for this question.

Manish said: (Dec 5, 2015)  
Square root of (4*s/a).

S = 0.2*Speed of overtaken vehicle + 6 (Base length of vehicle).

A = Acceleration.

Pavan said: (Jun 20, 2016)  
First change the overtaken vehicle speed into m/sec by multiply with .278 then apply

S = .7*- speed of overtaken vehicle + wheel base ( assume 6 meter)

Where a = acceleration

And T = Square root of (4* s/a)

Leo Basil George said: (Mar 24, 2017)  
Overtaking time T = [2(s1+s2)/a]^1/2.
Where s1 , s2 = .7v+l.
v - velocity of overtaken vehicle.
s1 -distance b/w overtaking and over taken vehicle.
s2- distance covered after overtaking.
l = wheel Base of vehicle ~ 6.1m.
a -overtaking acceleration =1.25 m/s2.
v = 30 * 5/18=8.33m/s
s1 = s2= .7 * 8.33+6.1 = 11.93m.

T = [2x(11.93+11.93)/1.25]^1/2.
= 6.17 seconds.

Anbarasan said: (Aug 29, 2017)  
FORMULAS:

OSD = 0.278Vbt + 0.278Vb T + 2(.2Vb +6) + VT

WHERE T = (4S/a)^0.5 and S = .2vb+6 a= acceleration= 1.25 Vb=30kmph V=50 KMPH

SOLUTION : S= .2 * 30 + 6 =12 T= (4*12/1.25)^.5= 6.197(ANSWER)
OSD= .278*30*2.5 + .278 *30*6.197+2(12)+50*6.197.

Gubendhiran said: (Dec 21, 2017)  
T=√[(4S)÷a].
S=6+(0.7Vb).
Vb=Design speed - 16 =50 - 16 =34 kmph,
S=6 + (.7*0.278*34) =12.61 m/s,
T=√[(4*12.61)÷1.25] = 6.35 seconds.
So option B was correct.

Umesh said: (Jul 24, 2019)  
Thanks @Gubendhiran.

Renis Gajipara said: (May 1, 2020)  
V= 50 km/ph = 13.88 m/s ....vehicle with design speed
Vb= 50 km/ph = 13.88 m/s....vehicle moving at a speed
a = 1.25 m/s^2.

Now,
S =0.7*VB +6 = 11.831 sec.
T = √(4S÷ a) = 6.15 sec.

Jahangir Abbas Sahil said: (Aug 28, 2021)  
V= 50 km/ph = 13.88 m/s -> vehicle with design speed.
Vb= 30 km/ph = 13.88 m/s -> vehicle moving at a speed.
a = 1.25 m/s^2.

Now,
S = 0.7*VB +6 = 11.831 sec.
T = √(4S÷ a) = 6.14 sec.

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