Quick math questions on acceleration and deceleration.

[Epidemic]

How fast will you go if you accelerate at 1G for 10 seconds?

Do I have this right? Would it be 2181.818182MPH? or 3200 feet per-second?

A bullet traveling at 3000 fps by the time it reaches the muzzle of the gun has traveled accelerated for 36 inches, how many G's does it experience?

A car stopping on an object like a brick wall at 60 mph decelerates at over a period of the metal crunching 36 to 40 inches: what can the occupant of said vehicle expect to encounter in g-force land.

If one tries to decelerate from the speed of light at 5 g's how long will it take?

[MTAtech]

If I remember from High School, the acceleration from gravity is 9.8m/sec. Thus, after 10 seconds an object would have a velocity of 98 meters per second after 10 seconds. A meter is roughly 3 ft., so, I get 294 ft/sec. or ~200 miles per hr.

I don’t think the bullet accelerates during its 36 inch voyage since there is no force acting upon it after the initial explosion. I would say that the actual acceleration is the first ˝ inch, making the G force about 25,000Gs.

It is difficult to calculate the car example as the deceleration is non-linear. However, I'll say ~100Gs – going from 60-0 in 30 inches.

The speed of light is 299,274 kilometers/sec. or 299,274,000 meters/sec. 5 Gs is (9.8 x 5)=49 meters/sec. So, I guess 70.69 days.

[Prexaspes]

Well the first quesions are simple math, 1G = 9.8 m/s²

The last one about decelerating from the speed of light...

It would take an infinite amount of time to decelerate from the speed of light since it would take an infinite amount of energy to get a massive object up to that speed, and when it got there it would have infinite mass and therefore infinte inertia.

edit : The bullet is acted upon until it reaches the end of the barrel since it is being pushed by expanding gas in the barrel, which is why a longer barrel yields a more accurate shot, not just rifling. The bullet will be traveling faster, giving it more inertia, making it's path harder to deflect. Especially true with a rifle's bullet, as the rifling adds rotational inertia to the equation.

edit #2 : I can't believe you guys let me get away with saying infinite weight! It's infinite mass.

[Theophylact]

(1) 320 feet/second. 1 G in the English system is 32 feet/second˛. Times 10 seconds, that's 320 feet/second. And after 10 seconds, you'll have traveled 1600 feet (d = ˝at˛, where d = the total distance and a = acceleration).

(2) At 1875 G, or 60,000 feet/second˛, a bullet would travel 3 feet in a hundredth of a second, and its speed would be 3000 feet/second. This is assuming constant acceleration. MTAtech's wrong; the explosion isn't like a sudden impact -- the expanding gases continue to accelerate the slug during its travel through the barrel. That's the main reason you have a barrel, other than to provide direction. The muzzle velocity will be greater with a rifle than with a stubby pistol for a given charge of powder.

(3) I haven't worked this one out, but you can see how it goes. At 60 miles per hour, that's 88 feet per second. You want to do this over a distance of about 3 feet.

(4) Forever. Now we're talking about relativity. Just as you can accelerate at at any rate for however long you like without ever reaching the speed of light, deceleration would also require an infinite amount of time. If you want to do the calculation, you'll have to choose some speed less than that of light; and if you want to keep relativity pretty much out of the calculations, keep the speed under 50% of c.

Dang, Prexaspes, you beat me to it; takes me too long to calculate.

[Epidemic]

wouldn't it be 9.8 M*S^2?
9.8 m/s^2 would be slower the longer it fell.

No, it's actually just m * s.

m/s^2 would give you some creepy inverse parabola. Not quite sure about that one.

Great, now my brain hurts.

Acceleration is the measure of the change in velocity over a specific distance and time. It ends up being an abstract number, which is why it's m/s^2. Right?

[Gomer]

Acceleration is m/s˛. It is not an abstract number. It is the change in velocity divided by the time over which the change occured.

Velocity is m/s.

Vf-Vi = ∆V -- m/s-m/s = m/s

∆V/t = a -- (m/s)/s = (m/s)(1/s) = m/s˛

[ShawnD1]

Originally posted by Epidemic

If one tries to decelerate from the speed of light at 5 g's how long will it take?

speed of light is something like 3x10^8 m/s I think.

t = Vf/A
t = (3x10^8)/(5 x 9.8)
t = 6122448 seconds which is 102040 minutes which is 1700 hours which is 70.8 days..... significant digits turns it into 71 days.

Not including relativity, it would take you 71 days to get to light speed