# Velocity after a certain acceleration.

sonhouse
Posers and Puzzles 17 Mar '09 06:13
1. sonhouse
Fast and Curious
17 Mar '09 06:13
I am reading a sci fi story by Michail McCollum called 'the Sails of Tau Ceti' and I am checking his math. At one point in the story, he has the intrepid hero's going out in a spacecraft and getting to "12 times the distance from pluto to the sun in 6 weeks". However, a couple pages on, he says the spacecraft reaches 4 percent of C in the same time. I think different. Anyone want to try the math to see how accurate that statement is? The craft is accelerating at a steady rate, except for short periods to jettison fuel tanks, so for all intents and purposes, it can be called a steady state acceleration.
So what do you say the craft is doing in terms of G's of accel, and what percent of C does it reach in that time. They have to get to about one light month away from Earth for the plot so after 6 weeks of accel, they coast then six weeks of decel. They spend 510 days coasting at top speed which gets them to 650 light hours from Earth according to McCollum. Will they get that far in 510 days after accel for the figures mentioned?
2. AThousandYoung
All My Soldiers...
18 Mar '09 05:534 edits
Originally posted by sonhouse
I am reading a sci fi story by Michail McCollum called 'the Sails of Tau Ceti' and I am checking his math. At one point in the story, he has the intrepid hero's going out in a spacecraft and getting to "12 times the distance from pluto to the sun in 6 weeks". However, a couple pages on, he says the spacecraft reaches 4 percent of C in the same time. I think o McCollum. Will they get that far in 510 days after accel for the figures mentioned?39.48
The average distance from Pluto to the Sun is 39.48 AU. An AU is 149,597,870,691 meters. Assuming he's talking about 12 times the average distance of Pluto from the Sun, the distance is 1795174448292 meters. To keep it simple, let's use 1.8 x 10^12 m.

Let's imagine the spacecraft moved the entire distance at 0.04c,or 1.2 x10^7 m/s. At this speed, it would take 150,000 seconds, or about 6 days.

So, at that speed it's easily possible. I guess we have to do the hard math then ๐

We know initial speed (0 m/s) and we know final speed at the halfway point (1.2 x 10^7 m/s). We know this equation works:

x = xi + 1/2(v+vi)t

Since initial speed and displacement are zero,

x = 1/2vt

1.8 x 10^12m = 1/2*(1.2*10^7 m/s)(t)

Now if we solve for t and get anything more than 3 weeks we know it doesn't work. I'm getting bored now though so I'll stop here.