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Motion in one dimension, Kinematics equations

  • #1

Homework Statement



Two spacecraft are 13,500m apart and moving directly toward each other. The first spacecraft has in initial velocity of 525 m/s and accelerates at a constant -15.5 m/s^2.
They want to dock, which means they have to arrive at the same position at the same time with zero velocity. (a) What should the initial velocity of the second be? (b) What should be it's constant acceleration?

SC1 Vi = 525 m/s
SC1 Vf = 0.0 m/s
SC1 a = -15.5 m/s^2
SC1 d = 8891.129032 m
SC1 t = 33.870967

SC2 Vi = unknown
SC2 Vf = 0.0 m/s
SC2 a = unknown
SC2 d = 4608.870968
SC2 t = I think it would be the same time as SC1?

Homework Equations



Vf = Vi + at
d = Vit + 1/2at^2
Vf^2 = Vi^2 + 2ad
d = 1/2(Vi + Vf)t

The Attempt at a Solution



I used (Vf^2 = Vi^2 + 2ad) and solved for d to find out the distance SC1 would travel until it's velocity reached 0 m/s

I then subtracted that from 13,500 to find d for SC2.

I used (Vf = Vi + at) to find t for SC1

Because the problem says they need to be there at the same time I thought I would use the time for SC1 for SC2 but using (d = 1/2(Vi + Vf)t) and solving for Vi gave me 9,217 m/s.... sounds bogus to me.

I know I need Two equations for Two unknown variables and solve them as one equation but I just don't know how to get there and put them together. This is my first physics class and I just need an example of what I need to do to get going. I tried every possible combination I could think of and I've realized that I clearly am misunderstanding something.

Thanks in advance for any and all help

-Chuck
 

Answers and Replies

  • #2
42
0
Your time for space craft 2 should be the same, then you could use d=1/2(Vi+Vf)t to solve for the initial velocity.
 

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