Problem 1 One key concept in transportation engineering is s

Problem 1 One key concept in transportation engineering is stopping sight distance. Stopping sight distance can be defined as the total distance required for a vehicle to stop. To calculate this distance, you have to account for two situations or variables: (1) brake reaction distance, and (2) braking distance: Stopping Sight Distance- Brake reaction distance +Braking distance Brake reaction distance is the distance traveled from the moment the driver recognizes a hazard on the road ahead of them and the moment he or she actually applies the brakes. This distance needs to account for the velocity at which the vehicle is traveling and the time it takes the driver to react to the hazard. Braking distance is the distance a vehicle travels between the driver applying the brakes and the car coming to a full stop The following problem statements refer to the brake reaction distance portion of the stopping sight distance calculation. (a) Which of the rectilinear kinematic equations is most appropriate for determining brake reaction distance? Justify your answer and list relevant assumptions. (b) The table below relates the reaction times in consideration of complexity and driver tate: Table 2 Reaction Times Considering C sity and Driver State Reaction Time 1.5s 3.0 s 2.5 s 2.5s s State C Alert Low Volume Road Low Two-Lane Primary Rural Road Fatigued Moderate Urban Arterial Alert Fatigued Fatigued Rural Freeway Low High Urban Freeway Calculate the brake reaction distance for each of the following situations. Assume that the vehicle is travelling at 50 miles per hour in each situation. (i) Low volume road, alert driver, low complexity [Answer: 110 ft] ( Two-lane primary rural road, fatigued, moderate complexity (ii Urban arterial, alert, high complexity (iv) Rural freeway, fatigued driver, low complexity [Answer: 183 ft (v)Urban freeway, fatigued, high complexity Relevant Resource:

Solution

Solution:-

(a) Answer :-

Kinematic equation

x = u t + (1/2) at²

This equation is used for braking reaction distance. Vehicle travelling at a constant velocity so acceleration a =0

Then braking reaction distance(x) = u t

Braking reaction distance depends on velocity of vehicle and perception reaction time.

(b)

(i) solution:-

Given:-

Velocity of vehicle = 50 miles/hour

                                   = 50 * 5280/3600

                                   =73.33 feet/second

where 1 mile = 5280 feet

Perception reaction time = 1.5 second

Brake reaction distance = vt = 73.33 *1.5 = 110 feet   Answer

(ii)

perception reaction time (t) = 3 seconds

brake reaction distance = 73.33 * 3 = 220 feet

(iii) when perception reaction time = 2.5 seconds

Brake reaction distance = 73.33 * 2.5 = 183 feet   Answer

(iv)

When perception reaction time = 2.5 seconds

Brake reaction distance = 73.33 * 2.5 = 183 feet         Answer

(v)   brake reaction distance = vt = 73.33 *3 = 220 feet   Answer