Highway safety engineers build soft barriers so that cars hitting them will slow down at a safe rate. A person wearing a seat belt can withstand and acceleration of -3.0 102 m/s2. How thick should barriers be to safely stop a car that hits a barrier at 106 km/h?
See:
http://www.jiskha.com/display.cgi?id=1284853293
Why did the car go to therapy?
Because it had some serious road rage issues!
But in all seriousness, let's calculate the thickness of the barrier needed to safely stop a car at 106 km/h.
First, let's convert the speed to m/s.
106 km/h is equal to (106000 m / 3600 s) = 29.4 m/s.
Now, let's use the equation for deceleration to find the necessary value:
v^2 = u^2 + 2as
Here, v is the final velocity (which is 0 m/s as the car comes to a stop), u is the initial velocity (which is 29.4 m/s), a is the deceleration (which is -3.0 × 10^2 m/s^2), and s is the distance (which we are solving for).
Rearranging the equation, we get:
s = (v^2 - u^2) / (2a)
Plugging in the values, we have:
s = (0^2 - 29.4^2) / (2 * -3.0 × 10^2)
s = -29.4^2 / -600
s ≈ 1.44 m
So, the barrier should be approximately 1.44 meters thick to safely stop a car traveling at 106 km/h.
Remember, safety first!
To determine the thickness of the barrier required to safely stop a car, we need to find the distance over which the car's velocity can be reduced at a safe rate.
First, let's convert the car's initial velocity from km/h to m/s:
106 km/h = 106 * (1000 m / 1 km) * (1 h / 3600 s) = 29.44 m/s
Now, we need to find the deceleration required to bring the car to a stop. Since both the initial velocity and final velocity are zero (the car stops), we can use the following kinematic equation:
v^2 = u^2 + 2as
where
- v = final velocity (0 m/s)
- u = initial velocity (29.44 m/s)
- a = deceleration
- s = distance
Rearranging the equation, we have:
a = (v^2 - u^2) / (2s)
Substituting the values, we have:
-3.0 * 10^2 m/s^2 = (0 - (29.44)^2) / (2s)
Simplifying and solving for s:
-3.0 * 10^2 m/s^2 = -864.4736 m^2 / (2s)
-3.0 * 10^2 m/s^2 * 2s = -864.4736 m^2
s = (-864.4736 m^2) / (-6.0 * 10^2 m/s^2)
s = 1.44079 m
Therefore, the thickness of the barrier should be at least 1.44 meters to safely stop the car.
To determine the thickness of the barriers required to safely stop a car, we need to understand a few concepts:
1. Conversion of units: We have the car's initial speed in kilometers per hour (km/h), but we need to convert it to meters per second (m/s) to match the unit of acceleration.
2. Using the kinematic equation: We can use the kinematic equation, which relates acceleration, initial velocity, final velocity, and displacement, to calculate the required deceleration and stopping distance.
Let's break down the problem step by step:
Step 1: Convert the initial speed from km/h to m/s.
- We know that 1 km/h is equal to 1000 m/3600 s.
- Therefore, 106 km/h is equal to (106 * 1000) m/3600 s, which simplifies to 29.44 m/s.
Step 2: Calculate the required deceleration.
- We know that acceleration (a) is given as -3.0 * 10^2 m/s^2.
- Since the car is decelerating, the acceleration is negative.
- Therefore, the required deceleration is -3.0 * 10^2 m/s^2.
Step 3: Determine the stopping distance.
- The stopping distance is the displacement covered by the car during deceleration.
- We can use the kinematic equation: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
- In this case, the final velocity (v) is 0 m/s since the car stops.
- The initial velocity (u) is 29.44 m/s (the converted speed from Step 1).
- We substitute these values into the equation: (0)^2 = (29.44)^2 + 2 * (-3.0 * 10^2) * s.
- Simplifying the equation, we find that s = (29.44^2) / (2 * 3.0 * 10^2).
Step 4: Calculate the required barrier thickness.
- The required barrier thickness is equal to the stopping distance (s).
- Therefore, the barrier thickness should be (29.44^2) / (2 * 3.0 * 10^2) meters.
Now you can plug the values into a calculator to determine the exact thickness of the barrier needed to safely stop the car.