Relative motion of ship

  • Thread starter madah12
  • Start date
  • #1
326
1

Homework Statement



At noon, ship A is 150 km west of ship B. Ship A is sailing east
at 35 km
h and ship B is sailing north at 25 km
h. How fast is
the distance between the ships changing at 4:00 P.M.?

Homework Equations





The Attempt at a Solution


I did it using geometry and related rates and got 21.5 which was the answer
but I when I try to do it using relative velocity it doesnt work:
I am taking the first ship initial location as the origin
rai=0
rbi=150i
ra=0+35ti
rb=150i+25tj
ra/b=a/g +ag/b
ra/b=0+35ti - 150i+25tj
ra/b=(35t-150)i + 25tj
dr(a/b) /dt = 35i+25j
and the magnitude is the way more than 21.5
I feel like I am doing something really wrong or setup the problem wrong.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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252
hi madah12! :smile:

suppose that a is stationary at (0,0), and b is at cosθti + sinθtj …

then using your method dr/dt = sinθti - cosθtj,

which is not the same as d|r|/dt, is it? :wink:
 
  • #3
326
1
oh yes I see d|r|/dt = 0 right cause |r|= 1 for all t? but then does the relative motion law holds for distances if not then I can't solve this problem using relative motion?
Edit
but we can still say that |s|= dD/dt = d ((25t)^2 + (150-35t)^2)^1/2 /dt right?
 
Last edited:
  • #4
tiny-tim
Science Advisor
Homework Helper
25,836
252
but we can still say that |s|= dD/dt = d ((25t)^2 + (150-35t)^2)^1/2 /dt right?

(have a square-root: √ and try using the X2 tag just above the Reply box :wink:)

right! :smile:
 

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