The Starship Enterprise (variable acceleration problem) part II

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SUMMARY

The discussion focuses on the variable acceleration problem related to the equation F = -be^(-a*v) = m dv/dt. The user, frankR, initially misapplied integration rules, particularly in evaluating the exponential function during integration. The correct approach involves evaluating the exponential function at the limits of integration, leading to the expression m/a(e^(a*v) - e^(a*vo)) = -b*t. This correction is crucial for accurately solving the problem and avoiding similar mistakes in future calculations.

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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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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).
 
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.
 
Last edited:

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