Question:
this is for driver's education safety information
Answers:
Why In the UK, car number plates are a diffrent colour?
2. convert to like terms
20 mph = 20 x 5280 feet per mile = 105600 feet per hour
105600 feet per hr x 12 inches per foot =1,267,200 inches/hr
1,267,200 in/hr ÷ 39 in. per meter = 32,492 meters per hr
32,492 meters/hr ÷ 60 minutes = 541.54 meters per min
541.54 meters/min÷ 60 seconds =9.025 meters per second
need to buy real accident occured at wofk site espicially petrochemical industry?
3. The velocity needed to be going at that is equal to 20 mph
about 9 meters per second which is obtained in 1 second
of free fall
4. Convert back to feet: (if the velocity was already constant at 9m/sec) but it is not .. it starts out as 0mph
(9 meters x 39 inches per meter)÷ 12 = # of feet fallen (29.33 ft)
Error.... I was wrong on the end of this calculation. I did not take into account the object starts out from 0 mph. This equation needed to be taken 1 step further to account for that. The actual height is 13+feet or approx 1 story. Hsueh's approach does include this final step and should win this one. It is nice the system allows for communication between people.
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It would only be one or two stories. Terminal velocity is 120mph, as in the fastest ANYTHING can fall given that it isnt powered.You can test the speeds by dropping anything that wont offer up much wind resistance from various heights (you need to measure the distance and time the fall). A basketball or something similar would be an ideal object to use.
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This is actually a really simple calculus problem. Or if your not in calculus, you'll need to know some basic distance formulas.I prefer to just derive them since it's much easier to understand.
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We are going to make the assumption that wind resistance is neglible.
We know gravity accelerates everything at a constant.
a = 32.2 ft/sec^2
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but acceleration is simply dv/dt
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dv/dt = 32.2 ft/sec^2
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dv = 32.2 ft/sec^2 dt
Integrate
v = 32.2 ft/sec^2 t + v(0)
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Assuming the car is not moving at t = 0, v(0) = 0
we end up with
v = 32.2 ft/sec^2 * t
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We want to know when the velocity will be 20 miles/hr (convert this to ft/sec since those are the units of gravity that I have memorized. 20 miles/hr * 5280ft/mile * 1hr/3600sec = 29.333 ft/sec
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Substituting this in for v, we can see how long it will take the car to reach 20 miles/hr.
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29.333 ft/sec = 32.2 ft/sec^2 t
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(29.333ft/sec)/(32.2 ft/sec^2) = t
So it will take
t = 0.910973084886128364 seconds to reach that speed (or less than a second to get to 20 miles per hour)
But how high off the ground does this need to be?
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We know that velocity v = ds/dt (Substituting into the above velocity equation)
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ds/dt = 32.2 ft/sec^2 * t
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ds = 32.2 ft/sec^2 * t dt
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Integrate where s represents distance. (assume positive down)
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s = 32.2 ft/sec^2 * (t^2)/2 + s(0)
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Again, assume that our zero is where we are going to begin our drop s(0) = 0
s = 32.2 ft/sec^2 * (t^2)/2
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s = 32.2 ft/sec^2 * (0.9109730^2)/2
Thanks to Rx_Mich who noticed that Yahoo actually cut off my numbers above when I was doing the substitution so it didn't make sense. I've shortened the number of significant digits so the values appear properly.
Dont you hate it when you are driving and....?
s = 13.3609 ft.
So you need to be 13.3609 ft off the ground so when you drop the car it will hit the ground at 20 mph, again this assumes zero wind resistance. (approximately one story high)
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Hope that helps.
I have a RV trailer and travel a lot with all the nuts on the road?
Simple right!
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I know this is easier if you just used algebratic distance formulas.
The formulas you would need are:
v = at + v(0)
(which I derived a = 32.2 ft/sec^2, and v(0) = 0 assume initial velocity of zero)
s = (1/2) at^2 + v(0) t + s(0)
(again I derived, a = 32.2 ft/sec^2, t calculated from above velocity formula, v(0) = 0, and s(0) = 0 assumed initial start position is zero)
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