Civil Engineering - Highway Engineering - Discussion
Discussion Forum : Highway Engineering - Section 6 (Q.No. 9)
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
Discussion:
10 comments Page 1 of 1.
Anila said:
2 years ago
Thanks everyone for explaining the answer.
(2)
Suhira said:
3 years ago
T= √(4S/a).
S= 0.7V + 6.
V = 8.33 m/s,
a = 1.25 m/s.
Thus, T=6.15 sec.
S= 0.7V + 6.
V = 8.33 m/s,
a = 1.25 m/s.
Thus, T=6.15 sec.
(5)
Jahangir Abbas Sahil said:
4 years ago
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.
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.
(1)
Renis Gajipara said:
5 years ago
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.
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.
(1)
Umesh said:
6 years ago
Thanks @Gubendhiran.
Gubendhiran said:
8 years ago
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.
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.
(2)
ANBARASAN said:
8 years ago
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.
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.
(2)
Leo Basil George said:
8 years ago
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.
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.
Pavan said:
9 years ago
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)
S = .7*- speed of overtaken vehicle + wheel base ( assume 6 meter)
Where a = acceleration
And T = Square root of (4* s/a)
(2)
Manish said:
10 years ago
Square root of (4*s/a).
S = 0.2*Speed of overtaken vehicle + 6 (Base length of vehicle).
A = Acceleration.
S = 0.2*Speed of overtaken vehicle + 6 (Base length of vehicle).
A = Acceleration.
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