A vehicle moving at 60 kmph on an ascending gradient of a highway has to come to a stop position to avoid collision with a stationary object. The ratio of lag to brake distance is 6:5 Considering total reaction of the driver as 2.5 seconds and the coefficient of longitudinal friction as 0.36, the value of ascending gradient (%) is

3.3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4.8

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

5.3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6.8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 4.8
Lag Distance
$= 0.278 V t$

$= 0.278 \times 60 \times 2.5$

$= 41.7 m$

Brake distance
$= \frac{5}{6} \times 41.7 = 34.75 m$

$[\frac{Lag distance}{Braking distance} = \frac{6}{5}]$

Braking distance
$= \frac{V^{2}}{254 (f + 0.01 n)}$

$\Rightarrow 34.75 = \frac{(60)^{2}}{254 (0.36 + 0.01 n)}$

$\Rightarrow 0.01 n = \frac{(60)^{2}}{254 \times 34.75} - 0.36$

$\Rightarrow n = 4.8 %$

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. A car is moving at a speed of

72 k m / h r

on a road having

2 %

upward gradient. The driver applies brakes when he sees an obstruction. If his reaction time is

1.5 seconds

, assuming that the coefficient of friction between the pavement and tyre as

0.15

, calculate the distance traversed before the car finally stops

Q. The coefficient of friction in the longitudinal direction of a highway is estimated as 0.396. The braking distance for a car moving at a speed of 65 km/hr is

A car travelling at a speed of 80 kmph is brought to a speed of 60 kmph within a distance of 30 m by applying brakes. On travelling further at a speed of 60 kmph, the driver of the car saw an overturned truck across and he applied the brakes again. The car stopped just close to the overturned vehicle. What was the distance (in kmph) between the overturned truck and the car when the driver first saw the overturned truck? The reaction time of the driver may be taken as 2 seconds.

An ascending gradient of 1 in 45 meets a descending gradient of 1 in 60. A summit curve has to be designed gradient for design speed of 80 km/h so as to provide a safe overtaking sight distance of 230 m. Estimate the length of the summit curve.

Join BYJU'S Learning Program

The correct option is B 4.8Lag Distance =0.278Vt =0.278×60×2.5 =41.7 m Brake distance =56×41.7=34.75 m [Lag distanceBraking distance =65] Braking distance =V2254(f+0.01n) ⇒34.75=(60)2254(0.36+0.01n) ⇒0.01n=(60)2254×34.75−0.36 ⇒n=4.8%