A vehicle is moving on a road oT slope +4% at a speed of 20 m/s. Consider the coefficient of rolling friction as 0.46 and acceleration due to gravity as 10 m/s2. On applying brakes to reach a speed of 10 m/s, the required braking distance (in m, round off to nearest integer) along the horizontal, is
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- A vehicle is moving down at a speed of 80 kph along an inclined surface (G = 2%). If the coefficient of friction is 0.33, compute the braking distance in meters. Round off to two decimal places.compute the braking distance (in meters). if a vehicle is moving down at a speed of 83 kph along an inclined surface (G = 2%). and the coefficient of friction is 0.33.Find the angle of banking for a highway curve of 90 m radius for cars traveling at 128 Km/hr, if the coefficient of friction between the tires and the road surface is 0.40. What is the rated speed of the road? (In rated speed, the friction force between the tires and the road is zero)
- A car is to be driven up a 7° incline road. Determine the automobiles braking distance from 30 m/s if the brakingdistance at 25 m/s speed is 45m when applied on a horizontal plane. Determine the braking distance going down a 5% incline. Assume the braking force is independent of the slope.A vehicle moving at a speed at a speed of 90 kph along an incline surface having aslope of 5%. If the coefficient of friction is 0.20, determine the braking distance.A moving car is traveling at 80 kph when the brakes are applied to it. The car stillmoves at a distance of 35 m, before it completely stops. Determine the coefficientof friction between the tires and the road surface.
- Situation 3. A road having a radius of 120m and has angle of 9.31 degrees from the horizontal. The center of gravity of the car is located 0.80m above the roadway and the distance between the two front wheels is 1.20m. If the car has a total weight of 15 KN, a. Compute the normal acceleration and the velocity of the car before overturning. b. If uk= 0.60, what is the velocity of the car without sliding? Please answer this with a complete solution and with a Free Body Diagram PLEASE. Thank you.For a car travelling with a speed of 50kph around a curve of radius 50m., Find the angle at which a road should be banked so that no friction is required.A car has a wheelbase of 250cm and a center of gravity that is 100cm behind thefront axle at a height of 60cm. If the car is traveling at 130 kph on a road with poor pavement thatis wet (μ = 0.6), determine the percentage of braking forces that should be allocated to the frontand rear brakes (by the vehicle’s braking system) to ensure that maximum braking forces aredeveloped.
- A vehicle weighing a 50 kN is moving at a constant speed around a circular curve. Neglecting the friction between the tires and the pavement and the centrifugal ratio (the ratio of the centrifugal force experience by the vehicle on the curve to its own weight) is 0.30. The degree of the curve is 5 degrees.a. Calculate the centrifugal force.b. Calculate the maximum speed the vehicle could move around the curve (in kph)c. If the skid resistance is 0.15, calculate the maximum super elevation that can be provided for the speed calculated from b.In a certain situation it was estimated that 24.9% of the braking force was applied to the rear brakes in order for the car to develop the maximum forces required stop the car. If the total braking force develop was 5565 N and the road is wet (u = 0.6) wheel base of 295 cm and a center of gravity 75 cm above the pavement road and 120 cm behind the front axle. What was the speed of the car in kph?TOPIC: Equation of Motion (NORMAL AND TANGENTIAL) Fn= Man Ft= Mat A road having a radius of 120m and has an angle of 9.31° from the horizontal. The center of gravity of the car is located 0.80m above the roadway and the distance between the two front wheels is 1.20m. If the car has a total weight of 15KN, Compute the normal acceleration and the velocity of the car before overturning. If uk= 0.60, what is the velocity of the car without sliding? PLEASE ANSWER WITH A COMPLETE AND DETAILED SOLUTIONS. THANK YOU