Traffic Engineering, 4th Edition
《交通工程》,第四版
Roess, R.P., Prassas, E.S., and McShane, W.R.
Roess,R.P.,Prassas,E.S.和 McShane,W.R.
Solutions to Problems in Chapter 2
第二章习题解答
Problem 2-1
习题 2-1
The driver continues to travel at 60 mi/h while taking his foot from the accelerator to the brake. Therefore, the vehicle will travel:
司机在将脚从油门移到刹车踏板的过程中,继续以 60 英里/小时的速度行驶。因此,车辆将行驶:
dr = 1.47 St = 1.47 * 60 * 3.5 = 308.7 ft before the driver’s foot hits the brake.
dr = 1.47 St = 1.47 * 60 * 3.5 = 308.7 英尺,然后司机的脚踩到刹车上。
Problem 2-2
习题 2-2
When the driver first sees the overturned truck, he will continue to move at 65 mi/h during his reaction time. During this time, the vehicle will travel 1.47*65*t feet, or 95.5t ft. Thus, the distance available for braking is 350 – 95.5t feet. This is the distance that is available for deceleration before the vehicle hits the overturned truck. The formula for braking distance is
当司机第一次看到倾覆的卡车时,他将在反应时间内继续以 65 英里/小时的速度行驶。在此期间,车辆将行驶 1.47*65*t 英尺,或 95.5t 英尺。因此,可用于制动的距离为 350 – 95.5t 英尺。这是车辆撞上倾覆的卡车前可用于减速的距离。制动距离的公式为:
In this case, the initial speed (Si) is 65 mi/h. The friction factor, F, is related to the deceleration rate, and is computed by dividing the deceleration rate by the deceleration rate due to gravity, or 10/32.2 = 0.31. The grade is level, i.e., G = 0. The braking distance is 350 – 95.5t. Therefore:
在这种情况下,初始速度(S)为 65 英里/小时。摩擦系数 F 与减速度率有关,其计算方法是将减速度率除以重力引起的减速度率,即 10/32.2 = 0.31。坡度为水平,即 G = 0。制动距离为 350 – 95.5t。因此:
This equation is solved for various values of t from 0.50 to 5.00 s. Note that at the point where the reaction distance becomes more than 350 ft, the final speed is a constant 60
此方程针对从 0.50 到 5.00 秒的不同 t 值求解。请注意,当反应距离超过 350 英尺时,最终速度为恒定的 60
mi/h, and the braking distance is essentially “0.”
英里/小时,制动距离基本上为“0”。