Calculate acceleration of a ship fired from a cannon

[hgducharme]

Homework Statement

Jules Verne [/B]In his novel From the Earth to the Moon (1866), Jules Verne describes a spaceship that is blasted out of a cannon, called the Columbiad, with a speed of 12,000 yards/s. The Columbiad is 900 ft long, but part of it is packed with powder, so the spaceship accelerates over a distance of only 700 ft. Estimate the acceleration experienced by the occupants of the spaceship during launch.

v₀ = 10,975.60976 m/s
a = ?
d = 213.4146341 m

Homework Equations

[/B]
1) x= x_0+v_0t+\frac{1}{2}at^2
2) v^2 = v_o^2 + 2a\Delta x

The Attempt at a Solution

I tried solving the second equation, but I got an insane number that isn't plausible. My main problem is what should I consider V₀, is it 0 or is it the 10,975 m/s? If I set it to 0, then I'm only calculating the acceleration from at rest to as soon as it leaves the cannon's barrel. Otherwise, if I set it to the 10,975 m/s, then I don't know the final velocity after accelerating for 213.4 m. Also, would the acceleration be considered as -9.8\frac{m}{s^2}?

 

[gneill]

The initial velocity is zero, the final velocity 12,000 yards/s. The distance over which the capsule accelerates is 700 ft.

Remember, this is fiction...

 

[NascentOxygen]

initial speed before the charge is lit is zero

final speed you know
distance over which it accelerates is given

find acceleration (the assumption is that acceleration is uniform)

 

[hgducharme]

The initial velocity is zero, the final velocity 12,000 yards/s. The distance over which the capsule accelerates is 700 ft.

Remember, this is fiction...

Haha, yes I picked up on that. They would have the speed to get past LEO after 213m.

So are you saying the final velocity is 12,000 yards/s AFTER the 213m uniformed acceleration?

 

[gneill]

Haha, yes I picked up on that. They would have the speed to get past LEO after 213m.

So are you saying the final velocity is 12,000 yards/s AFTER the 213m uniformed acceleration?

Yes, of course. That's how canon's work. The acceleration takes place while the projectile is within the barrel.

 

[hgducharme]

Ok, I still can't seem to figure out what I'm doing wrong. Here's my work:

v^2 = v_0^2 + 2a\Delta x Solving for a gives me:

\frac{v^2 - v_0^2}{2\Delta x} = a Plug in my values:

\frac{(12,000 yds/s)^2}{2(233.34 yds)} = a Simplify:

\frac{144,000 yds/s}{466.67 yds} = a

308.57 s = a

Some how I end up with only seconds as my unit. I assume I'm using the wrong formula?

Edit: Ahhhhh, nevermind. I forgot to square my units in the numerator!

 

[gneill]

$\left(\frac{yd}{s}\right)^2 = \frac{yd^2}{s^2}$ right?

And check the order of magnitude for your yards squared. You've lost several.

 

[hgducharme]

Yes, I just caught that, thank you