Solving Motion Equations: Two Spacecrafts 13,500 m Apart

[r_swayze]

I don't know why but I am having a little problem with this problem here:

Two spacecraft are 13,500 m apart and moving directly toward each other. The first spacecraft has velocity 525 m/s and accelerates at a constant −15.5 m/s2. 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 spacecraft be? (b) What should be its constant acceleration?

Here is my work:

used vf = vi + at to get 33.87 s for time for the 1st aircraft

used displacement = 1/2(vi + vf)t to get 8891 m

then 13500m - 8891m = 4609 m for distance traveled for the 2nd aircraft

used displacement = 1/2(vi + vf)t to get 272 m for the vi of the 2nd aircraft (answer to a)

then used vf = vi + at to get -8.04 m/s^2 for acceleration of the 2nd aircraft (answer to b)

if you plug these numbers in and add up both equations then they equal to 13500 m but somehow I am wrong? Am I using the wrong formulas or have made an err in my math? can somebody please help?

 

[Zaphys]

You should forget the eq. you use to calculate displacements

d=1/2·(Vi+Vf)t

Because is almost never working just because it's not true in general.

I didn't look into all your math" but I can ensure you that using that formulae you´ll do wrong.

 

[tms]

You should forget the eq. you use to calculate displacements

d=1/2·(Vi+Vf)t

Because is almost never working just because it's not true in general.

Assuming motion in one dimension, constant acceleration, and an initial position at $$x = 0$$, all of which conditions hold for this problem, the above equation is perfectly good.

 

[tms]

if you plug these numbers in and add up both equations then they equal to 13500 m but somehow I am wrong? Am I using the wrong formulas or have made an err in my math? can somebody please help?

Why do you say you are wrong?

 

[Jebus_Chris]

Lets draw a picture.
o₁-----------x-------------0₂
Velocity is positive to the right as shown in the question. Because of that velocity and acceleration of o₂ should be the opposite sign of o₁.

a<--- o₁ --->v
v<--- o₂ --->a

What you have is correct in magnitude but not direction, the velocity should be negative and the acceleration should be positive.

 

[Zaphys]

Sorry tms and swayze I was just thinking wrong... :S

never mind swayze you're all right in that sense.

 

[MikWillis]

I know it's really late, but just in case you wanted to know (a) would be -272 and (b) would be 8.04. You just had the signs mixed up. I have the same online homework.