Are pvc car seats safe?

My driver and passenger seats are aweful, theyre ripped, stained, the whole 9 yards (it is a 10 year old car after all), so i was thinking of getting the racing style ones with the 4 point harness, and i found pretty cheap ones on ebay, its 2 seats and 2 harnesses for about $300 with free shipping. the only thing i noticed is theyre pvc seats, i dont know what this means and have never heard of it, but seat belts can be the decision making factor if god forbid you get into an accident so i was curious if pvc seats are of good quality and safe overall, i imagine they wouldnt sell them if they werent safe but i could be wrong. thanks

That looks like a standard fall arrest harness. Don't worry about how you will look; they are uncomfortable enough when properly fitted that you won't give it a thought. Almost everybody leaves the leg straps too loose until they hang by the harness (we hang by it during rescue practice), then they decide to tighten the leg straps!

Source(s):

Do more tower work than I care to... part of the job.

A car is traveling at 25 m/s when it runs off the road and hits a utility pole. The car stops instantly, but the driver continues to move forward at 25 m/s. The airbag starts from rest with a constant acceleration from a distance of 50 cm away from the driver and makes contact with him in 9 ms. What is the acceleration of the airbag?

I got this as the answer a = 1234.568 m/s, but its wrong.

a foot is really pushing it, im sure youll be fine but like mokire said all it takes is one drunk, or carless driver to clip you when youre that far out. what i usually do is ride up onto the curb until i cant see the street in my passenger side mirror, that tells me im about 5 or 6 inches from the curb and am good to park

Let x be the number of passenger cars
Let y be the number of trucks

x + y = 108
0.25(2)x + 0.25 (5)y = 72

x + y = 108
0.50x + 1.25y = 72
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I'm going to solve this by substitution (but first, I'll get x by itself in the first equation, so I can sub it into the second equation and solve for y (trucks).

x + y - y = 108 - y <-- get x by itself by subtracting y both sides/
x = 108 - y
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Now I can substitute this equation for the x variable in the second equation and solve for the number of trucks...

0.50x + 1.25y = 72

0.50(108 - y) + 1.25y = 72

54- 0.50y + 1.25y = 72 <--- combine like terms (add -0.5y and 1.25)

54 + 0.75y = 72 <-- now solve for y by first subtracting 54 both sides.

54 + 0.75y - 54 = 72 - 54

0.75y = 18 <-- now divide both sides by 0.75 to get y.

0.75y/0.75 = 18/0.75

y = 24
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Since I assigned y to be the number of trucks, therefore, the number of trucks is 24.