Question

A Federation starship ( 1.80×10^6kg ) uses its tractor beam to pull a shuttlecraft ( 2.20×10^4kg ) aboard from a distance of 19.0km away. The tractor beam exerts a constant force of 4.0*10^4 N on the shuttlecraft. Both spacecraft are initially at rest.

How far does the starship move as it pulls the shuttlecraft aboard?

Answer 1

Here is my attempt at the solution. (4.0*10^4N)/(2.20*10^4kg) = a(shuttlecraft)
Use a to find final velocity with equation vf^2 = vi^2 + 2a*x -> vf = 8.312 m/s
Now that vf is found I used vf to find the time taken for the shuttlecraft to board the federation ship with the equation vf = vi + at -> t = 4.572s
I then did 4.4*10^4/1.8*10^6 = a(federation starship) = .022 m/s^2.
then I finally used the equation xf = xi +vit + (1/2)at^2 to compute the final position of the starship ang got .23 m which is the wrong answer.

Answer 2

What exactly is the question asking you to solve for?

Answer 3

the length the starship travels due to the force the tracktor beam pulls on the shuttlecraft.

Answer 4

ok I appreciate it.

Answer 5

This will take a minute to get it all typed out....but I came with a very different answer.

Answer 6

Haha I thought the 19km was 19 m. thanks a lot.

Answer 7

Its ok, I got the answer. 229.9m

Answer 8

I hate latex..makes for silly mistakes once you type it all out. Here it is redone...your answer is pretty close to mine:

\[F=4.00x10^4N\]\[m_{starship}=1.80×10^6kg\]\[m_{shuttle}=2.20×10^4kg\]\[x=1.90x10^4m\]
\[a_{starship}=\frac{4x10^4N}{1.8x10^6kg}=0.022m/s^2=2.22x10^{−2}m/s^2\]
\[a_{shuttle}=\frac{4x10^4N}{2.2x10^4kg}=1.82m/s^2\]
Solve for the time at which the ships (given their individual accelerations) meet:
\[1.90x10^4m=\frac{1}{2}(2.22x10^{−2}m/s^2+1.82m/s^2)(143.6s)^2\]\[t=143.6s\]
\[\Delta X_{starship}=\frac{1}{2}(2.22x10^{-2}m/s^2)(143.6s)^2=226.98m=227m\]

Answer 9

Considering your force was only given to 2 significant figures the answer would end up being simply 230m anyway