Question

Drivers with an average of 20/40 vision travel at 55 mph in the curb lane of a freeway, where exit ramps are designed for 25 mph. What should be the minimum distance of signs with 6-in. letters placed ahead of the exit?

The following information may be useful:
Perception-reaction time 2.5 seconds;
Deceleration rate - 5 ft/sec2:
Road level-1% (downgrade);
Drivers with 20/20 vision can read signs at 60 ft per inch of letter height. .

Answer 1

The sign board must be placed 573 ft ahead of the exit.

Step-by-step explanation:

Distance needed for reducing the speed from 55 mph to 25 mph is given as

`d=(v_f^2-v_i^2)/(30 * ((a)/(g)-G))`

Here

• 
  `v_f` is the velocity at the end which is 25 mph
• 
  `v_i` is the velocity at the start which is 55 mph
• a is the rate of deceleration which is -5 ft/s^2
• G is the Road grade which is 1% or 0.01
• g is the gravitational acceleration whose value is 32.2 ft/s^2

`d=(v_f^2-v_i^2)/(2 * ((a)/(g)-G))\\d=(25^2-55^2)/(30 * ((-5)/(32.2)-0.01))\\d=551 ft`

Now the perception time is 2.5 second, 20/20 vision person can read 6 inch letters from 60 x 6 ft.

For 20/40 vision person can read 6 inch letters from 30 x 6 ft=180 ft.

SSD is
`d=(55 * 5280 * 2.5)/(60 * 60)\\d=202 ft`

So the minimum distance is given as

`Minimum distance=551+202-180 ft\\Minimum distance=573 ft\\`

So the sign board must be placed 573 ft ahead of the exit.