Highway safety engineers build soft barriers so that cars hitting them will slow down at a soft rate. A person wearing a safety belt can withstand an acceleration of 300m/s^2.
How thick should barriers be to safely stop a car that hits the barriers at 92.8km/h?
Answer in units of m.
We can start by converting the car's speed from km/h to m/s.
92.8 km/h = 92.8 * (1000/3600) m/s = 25.78 m/s
To find the minimum distance (thickness of the barrier) required to safely stop the car, we need to calculate the deceleration (negative acceleration) required to bring the car to a stop.
The equation to calculate the deceleration (acceleration) is:
V^2 = U^2 + 2aS
where V is the final velocity (0 m/s), U is the initial velocity (25.78 m/s), a is the acceleration (deceleration), and S is the distance (thickness of the barrier).
Plugging in the values, the equation becomes:
0 = (25.78)^2 + 2(-a)S
Rearranging the equation, we get:
aS = (25.78)^2
Since we want to find the thickness of the barrier (S), we can divide both sides of the equation by a:
S = (25.78)^2 / -a
Given that a person wearing a safety belt can withstand an acceleration of 300 m/s^2, we can substitute this value into the equation:
S = (25.78)^2 / -(300)
Calculating this, we get:
S ≈ 0.704 m
Therefore, the minimum thickness required for the barrier to safely stop the car is approximately 0.704 meters.
To determine the thickness of the barrier needed to safely stop a car, we first need to calculate the deceleration experienced by the car. We can then use this deceleration and the initial speed of the car to find the appropriate thickness of the barrier.
1. Convert the initial speed of the car from km/h to m/s:
92.8 km/h = 92.8 * 1000 m / 3600 s = 25.78 m/s
2. Calculate the deceleration:
We can use the equation of motion v^2 = u^2 + 2as, where v = final velocity, u = initial velocity, a = acceleration/deceleration, and s = distance.
Since we want to decelerate from 25.78 m/s to a stop (0 m/s) and the acceleration is negative (deceleration), we have:
0^2 = 25.78^2 + 2 * a * s
Simplifying, we get:
0 = 664.6084 + 2as
-2as = 664.6084
as = -332.3042
3. Convert the deceleration to positive value:
Since the deceleration is given as -332.3042 m/s^2, we need to consider the absolute value of it for this calculation. So the deceleration becomes 332.3042 m/s^2.
4. Use the maximum acceleration a person wearing a safety belt can withstand to find the thickness of the barrier:
To stop the car at a soft rate and ensure the person inside is safe, the deceleration should not exceed 300 m/s^2.
Therefore, we equate the deceleration of the car to 300 m/s^2:
332.3042 m/s^2 = 300 m/s^2
5. Calculate the distance covered during deceleration:
Again, using the equation of motion v^2 = u^2 + 2as, but this time the final velocity (v) is 0 m/s, the initial velocity (u) is 25.78 m/s, and the acceleration (a) is -332.3042 m/s^2. We need to find the distance (s).
0^2 = 25.78^2 + 2 * (-332.3042) * s
Simplifying, we get:
0 = 664.6084 - 664.6082s
664.6082s = 664.6084
s = 1 m (approximately)
Therefore, the thickness of the barrier should be approximately 1 meter to safely stop a car that hits it at a speed of 92.8 km/h.