way safety engineers build soft barriers

so that cars hitting them will slow down at a
safe rate. A person wearing a safety belt can
withstand an acceleration of 300 m/s
2
.
How thick should barriers be to safely stop
a car that hits the barriers at 81.7 km/h?
Answer in units of m

v=81.7 km/h =81700/3600=22.7 m/s

d= v²/2a=22.7²/300=1.72 m

To determine the thickness of the soft barriers required to safely stop a car, we need to calculate the distance the car will travel while decelerating from 81.7 km/h to rest.

First, let's convert the initial velocity of the car from km/h to m/s:
81.7 km/h = 81.7 * (1000 m/1 km) * (1/3600 h/1 s) ≈ 22.7 m/s

Next, we need to determine the deceleration of the car. Since the car hits the soft barrier and slows down, the deceleration is equal to the negative acceleration the person wearing a safety belt can withstand, which is given as -300 m/s^2.

Using the equation of motion: v^2 = u^2 + 2as, where:
v = final velocity (0 m/s as the car comes to rest)
u = initial velocity (22.7 m/s)
a = acceleration/deceleration (-300 m/s^2)
s = distance traveled

Rearranging the equation, we get:
s = (v^2 - u^2) / (2a)

Substituting the values into the equation:
s = (0^2 - 22.7^2) / (2 * -300)

Calculating this:
s ≈ (-516.29) / (-600) ≈ 0.86 m

Therefore, the thickness of the soft barriers should be approximately 0.86 meters to safely stop a car traveling at 81.7 km/h.