A person wearing a safety belt can withstand an acceleration of -3.0E2 m/s2. How thick should barriers be to safely stop a car that hits a barrier at 97 km/h?

First of all, do the conversion

97 km/h = 26.9 m/s

If deceleration occurs over a distance X (the "barrier thickness")

Vo = sqrt(2 a X) ; therefore

X = Vo^2/(2 a) = (26.9)^2/(2*300) = 1.21 m

To determine the thickness of barriers required to safely stop a car, we need to understand the concept of deceleration and use the provided information about the person's ability to withstand acceleration.

First, let's convert the speed of the car from km/h to m/s. We know that 1 km/h is equal to 0.2778 m/s. So, the speed of the car is:

97 km/h * 0.2778 m/s = 26.9466 m/s

Now, we need to calculate the deceleration (negative acceleration) required to stop the car. We can use the following equation:

final velocity^2 = initial velocity^2 + 2 * acceleration * distance.

Since the final velocity is 0 m/s (as the car needs to stop), the initial velocity is 26.9466 m/s, and the acceleration is unknown (denoted as "-a" to represent deceleration), we can rewrite the equation as:

0 = (26.9466 m/s)^2 + 2 * (-a) * distance.

Simplifying this equation, we have:

a * distance = (26.9466 m/s)^2.

Since we are given that a person wearing a safety belt can withstand an acceleration of -3.0E2 m/s^2 (negative sign indicates deceleration), we substitute this value into the equation:

(-3.0E2 m/s^2) * distance = (26.9466 m/s)^2.

Now we can solve for the distance. Rearranging the equation:

distance = (26.9466 m/s)^2 / (-3.0E2 m/s^2).

Calculating this value:

distance = (26.9466 m/s)^2 / (-300 m/s^2) = 2.391 m (rounded to 3 decimal places).

Therefore, the required thickness of the barriers to safely stop a car traveling at 97 km/h is approximately 2.391 meters.