The Starship Enterprise (variable acceleration problem) part II

  • Thread starter frankR
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  • #1
frankR
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Okay what am I doing wrong? This is the way I've been doing math for the last two years. This is annoying me. Unless I've been doing everything wrong the last two years, I feel this is correct. I realize it's most likely wrong. Someone please explain to me what I am doing wrong and more important why.

F = -be^(-a*v) = m dv/dt, a and b are constants.

m [inte]vov e^(a*v) dv = -b [inte]to=0t dt

m/a e^(a(v - vo)) = -b*t

ln[e^(a(v - vo))] = ln[-abt/m]

a(v - vo) = ln[-abt/m]

v(t) = 1/a ln[-abt/m] + vo

dx/dt = v(t) = 1/a ln[-abt/m] + vo

[inte]xo=ox dx = [inte]to=ot1/a ln[-abt/m] + vodt

x(t) = t/a[ln(-a*b*t/m) + a*vo -1]
 
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Answers and Replies

  • #2
quantumdude
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Originally posted by frankR
m [inte]vov e^(a*v) dv = -b [inte]to=0t dt

m/a e^(a(v - vo)) = -b*t

Your mistake is in the last line here. When you do the integral, you have to evaluate exp(av) at v and at v0 and subtract, to get:

exp(av)-exp(av0).

This does not equal:

exp(a(v-v0)),

which is what you have. Incidentally, this is the same basic mistake that I pointed out in "Part I" of this problem, except there you did it with the inverse function (natural log), when you used the invalid rule:

ln(a+b)=ln(a)+ln(b).
 
  • #3
frankR
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m/a e^(a*v)|vov = -b*t

m/a(e^(a*v) - e^(a*vo) = -b*t

Okay now that makes sense.

Thanks.

Edit: Hopefully I won't make that mistake again.
 
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