Highway safety engineers build soft barriers such as the one shown in Figure 3-21, so that cars hitting them will slow down at a safe rate. A person wearing a safety belt can withstand an acceleration of -3.0 102 m/s2. How thick should barriers be to safely stop a car that hits a barrier at 121 km/h?
first 121 km/h (1000m/1km)(1h/3600s) = 33.6 m/s
vf2 = vi2 + 2a(&Delta x)
&Delta x is what we want to solve for, it tells us the distance which we will compress the barrier so it needs to be atleast that thick
a is the max acceleration given which I don't really understand from what you've written (-3.0 102?)
vf is 0 because the car will stop
vi is 33.6 m/s
solving for &Delta x we get
[vf2 - vi2] / 2a = &Delta x
To find the required thickness of the barrier to safely stop a car, we need to consider the deceleration experienced by the car.
Given:
Acceleration tolerance (a): -3.0 × 10^2 m/s^2
Initial velocity of the car (v): 121 km/h
First, let's convert the initial velocity from km/h to m/s:
1 km/h = 1000 m/3600 s
Therefore, the initial velocity is:
v = 121 km/h × (1000 m/3600 s) = 33.61 m/s (rounded to two decimal places)
The final velocity of the car when it comes to a stop is 0 m/s.
We can use the following equation of motion to find the required distance (d) for the car to stop:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s)
u = initial velocity (33.61 m/s)
a = deceleration (unknown)
s = distance covered (unknown)
Rearranging the equation, we have:
s = (v^2 - u^2) / (2a)
Substituting the given values into the equation:
s = (0 - (33.61)^2) / (2 × -3.0 × 10^2)
Calculating the value of s:
s = (0 - 1128.3121) / -600
s ≈ 1.88 m (rounded to two decimal places)
Therefore, the thickness of the barrier should be approximately 1.88 meters to safely stop the car that hits it.
To determine the thickness of the barrier required to safely stop a car, we need to calculate the deceleration experienced by the car when it hits the barrier and then find the thickness that can provide this deceleration.
Let's start by converting the car's initial velocity from km/h to m/s:
121 km/h * (1000 m/1 km) * (1 h/3600 s) = 33.61 m/s
Now, we need to calculate the deceleration (negative acceleration) experienced by the car. We can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s since the car comes to a stop)
u = initial velocity (33.61 m/s)
a = acceleration (deceleration)
s = distance covered (unknown, the thickness of the barrier)
Rearranging the equation to solve for acceleration (deceleration):
a = (v^2 - u^2) / (2s)
Substituting the known values:
a = (0 - (33.61 m/s)^2) / (2s)
a = -33.61^2 / (2s)
a = -1130.3121 / s
We are given that a person wearing a safety belt can withstand an acceleration of -3.0 * 10^2 m/s^2. So, we can set up an equation to find the required thickness (s) that can provide this acceleration:
-3.0 * 10^2 = -1130.3121 / s
Multiplying both sides of the equation by s:
s * (-3.0 * 10^2) = -1130.3121
Solving for s:
s = -1130.3121 / (-3.0 * 10^2)
s ≈ 3.767 meters
Therefore, the thickness of the barrier should be approximately 3.767 meters to safely stop a car hitting it at 121 km/h.