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 an acceleration of -300 m/s². How think should barriers be to safely stop a car that hits a barrier at 110 km/h?
Vf^2=Vi^2-2ad
solve for d, that is the thickness of the barriers.
I had assumed as much but I couldn't find the appropriate formula in the chapter given...
While I'm sure that'll get me the correct answer I get the feeling that I was expected to use another formula.
Anything else come to mind?
It would help if you give the context.
Is your chapter on work and energy?
KE = kinetic energy = (1/2)mVi²
Work done
= force * distance
= ma * distance
Equate KE to Work done:
(1/2)mVi² = maD
or D=(1/2)Vi²/a
(Same result as what Mr. Pursley gave)
To calculate the thickness of the barrier required to safely stop a car, we need to determine the deceleration (negative acceleration) the car experiences when it hits the barrier. We know that a person wearing a seat belt can withstand an acceleration of -300 m/s².
First, we need to convert the speed of the car from km/h to m/s. We can do this by dividing the speed in km/h by 3.6:
110 km/h = (110/3.6) m/s ≈ 30.56 m/s
Now, let's calculate the deceleration using the formula:
v² = u² + 2as
Here, v is the final velocity (0 m/s because the car should come to a stop), u is the initial velocity (30.56 m/s), a is the deceleration, and s is the distance traveled (unknown, the thickness of the barrier).
0² = (30.56)² + 2a(s)
0 = 934.9936 + 2as
2as = -934.9936 (since final velocity is 0)
a = -934.9936 / (2s)
Now, we can substitute the maximum acceleration (-300 m/s²) and solve the equation for s (thickness of the barrier):
-300 = -934.9936 / (2s)
Multiplying both sides by 2s:
-600s = -934.9936
Dividing both sides by -600:
s ≈ -934.9936 / -600
s ≈ 1.558 m
Therefore, the barrier should be approximately 1.558 m thick to safely stop a car that hits it at 110 km/h.