If the design speed is 80 kmph, perception reaction time 3 seconds and coefficient of friction is 0.5, the safe stopping sight distance is

117.11

Open in App

Solution

The correct option is A 117.11
Stopping sight distance,

$SSD = (0.278 V t) + \frac{V^{2}}{254 f}$

Here,
V = 80 km/hr
Reaction time, t = 3 sec
Coefficient of friction, f = 0.50

$∴ SSD = (0.278 \times 80 \times 3) + \frac{{(80)}^{2}}{254 \times 0.50}$

$= 66.72 + 50.39 = 117.11 m$

Suggest Corrections

Similar questions

Q. What will be the non-passing sight distance (in m) on highway for a design speed of 80 kmph when its ascending gradient is 3%? Assume brake efficiency as 50%, coefficient of friction as 0.7 and reaction time of 2.5 seconds.

Q. The safe stopping distance for a design speed of 50 kmph for a two lane road with coefficient of friction of 0.37 in m is

Q. The design speed on a road is 60 kmph. Assuming the driver reaction time of 2.5 seconds and coefficient of friction of pavement surface as 0.35, the required stopping distance for two-way traffic on a single lane road is

The perception- reaction time for a vehicle travelling at 90 km/h, given the coefficient of longitudinal friction of 0.35 and the stopping sight distance of 170 m (assume g = $9.81 m / s^{2}$ , is_______seconds

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

If the design speed is 80 kmph, perception reaction time 3 seconds and coefficient of friction is 0.5, the safe stopping sight distance is 117.11

The correct option is A 117.11Stopping sight distance, SSD=(0.278Vt)+V2254f Here, V = 80 km/hr Reaction time, t = 3 sec Coefficient of friction, f = 0.50 ∴SSD=(0.278×80×3)+(80)2254×0.50 =66.72+50.39=117.11 m

If the design speed is 80 kmph, perception reaction time 3 seconds and coefficient of friction is 0.5, the safe stopping sight distance is

117.11

The correct option is A 117.11
Stopping sight distance,

$SSD = (0.278 V t) + \frac{V^{2}}{254 f}$

Here,
V = 80 km/hr
Reaction time, t = 3 sec
Coefficient of friction, f = 0.50

$∴ SSD = (0.278 \times 80 \times 3) + \frac{{(80)}^{2}}{254 \times 0.50}$

$= 66.72 + 50.39 = 117.11 m$