2 A vehicle travelling at a speed of 40 mph is at a distance

2. A vehicle travelling at a speed of 40 mph is at a distance of 200 ft from an intersection when the signal turned yellow. The width of the intersection is 50 ft and the length of the vehicle is 19 ft. If the duration of the yellow phase of the signal is 4 s, which of the following situation(s) is/are practically possible? Assume 1.5 s of reaction time. (a) The driver travels at the constant speed without stopping (b) The driver applies brake and stops at the stop-line before the intersection. (c) The driver accelerates to cross the intersection before the signal turns red. In this case what is the rate of acceleration needed?

Solution

initial speed of vehicle = 40 mph=17.8 m/s

distance from intersection=200 ft=60.96 m

width of intersection=50 ft=15.24 m

reaction time=1.5 s

duration of yellow phase = 4 seconds

length of vehicle=19 ft=5.79 m

a) if the vehicle continues with constant speed:

total distance to be travelled to cross intersection safely=60.96+15.24+5.79=81.99 m

time required to travel this distnace with speed of 17.8 m/s = 81.99/17.8=4.6 seconds

Since the yellow phase duraton is 4 seconds , travelling with same speed without stopping is not practical

b)distance travelled in reaction time = 17.8*1.5=26.7 m

remaining distance to be travelled =60.96-26.7=34.26 m

time available to stop=4-1.5=2.5 seconds

Let the deceleration required of the car to stop before the intersetion be d

d=17.8/2.5=7.12 m/s²

A car cannot decelerate at such a high rate. So, this option is also not practical

c) Distance travelled in 1.5 seconds=26.7 m

remaining distance to be travelled = 60.96+15.24+5.79-26.7=55.29 m

time in which this distance has to be covered=2.5 seconds

let the acceleration required be a

55.29=17.8*2.5 + 0.5*a*2.5²

a=3.45 m/s²

This seems a reasonable number. Therefore, this option is practical

The rate of acceleration needed is 3.45 m/s²