Question

A test of a driver’s perception/reaction time was conducted on a special test track in a rural area. The friction factor f = 0.6. The driver’s initial speed is 55 MPH. The track is on level ground.
a) When a driver is not texting on his smart phone, a stop can be made just in time to avoid hitting an unexpected object that is first visible 520 feet ahead.
b) When a driver is texting on his smart phone, but with all other conditions exactly the same, the driver is distracted and fails to stop and hits the object at a speed of 35 MPH. Determine the driver’s perception/reaction time for both situations, for (a) and for (b).

Answer 1

Answer:

a) tD = 4.36 sec

b) tD = 5.2 sec

Explanation:

a)

V = 0

work-done = change in kinetic energy

f.m.g(d) = 1/2m(55^2 - v^2)

divided both sides by m

1/2(55^2 - v^2) = f.g(d)

1/2(55 * 1.467)^2 ft^2/sec^2 = 0.6 * (9.81 * 3.28) * d

3255.03 = 19.306d

d = 3255.03 / 19.306

d = 168.6 ft

now D = 520 - 168.6 = 351.4

therefore reaction/perception time = tD

tD = 351.4 / ( 55 * 1.467)

tD = 4.36 sec

b)

Also here, V = 35 mph

so

1/2(55^2 - 35^2)*(1.467)^2 ft^2/sec^2) = 0.6 * (9.81 * 3.28) * d

1936.88 = 19.306d

d = 1936.88 / 19.306

d = 100.325ft

also D = 520ft - 100.325ft = 419.675

so tD = 419.675 / ( 55 * 1.467)

tD = 5.2 sec