Question

The stopping sight distance (SSD) for a level highway is 140 m for the design speed of 90 km/h. The acceleration due to gravity and deceleration rate are 9.81 m/s\(^2\) and 3.5 m/s\(^2\), respectively. The perception/reaction time (in s, round off to two decimal places) used in the SSD calculation is \(\underline{\hspace{2cm}}\).

Hint:

The stopping sight distance (SSD) includes the perception/reaction time and the braking distance, which is affected by the speed and deceleration rate.

Answer 1

Correct Answer: 1.9

The stopping sight distance (SSD) is given by the formula: \[ SSD = v_0 \times t + \frac{v_0^2}{2a}, \] where:
- \( v_0 = 90 \, \text{km/h} = 25 \, \text{m/s} \) is the design speed,
- \( a = 3.5 \, \text{m/s}^2 \) is the deceleration rate,
- \( t \) is the perception/reaction time.
Rearranging the formula to solve for \( t \): \[ t = \frac{SSD - \frac{v_0^2}{2a}}{v_0}. \] Substitute the given values: \[ t = \frac{140 - \frac{25^2}{2 \times 3.5}}{25} = \frac{140 - 89.29}{25} = \frac{50.71}{25} = 2.03 \, \text{s}. \] Thus, the perception/reaction time is \( \boxed{2.03} \, \text{s} \).