# The Starship Enterprise (variable acceleration problem)

The Starship Enterprise initially moving with speed vo hits an intergalctic metero shower and expreiences a deceleration force given by:

F = -b*e^([alpha]*v) where b and [alpha] are constants. The star ships mass is m.

a) Determine v(t).
b) Determine the time required for the Enterprise to stop.
c) Show that x(t) is given by: (a really ugly function I don't want to type)

I've solved it, however my x(t) function isn't like what is given. I certain that my math and physics is right. If someone could do the problem, I'm curious to know what you get.

For a)

Finding v(t)

F = -b*e^([alpha]*v) = m dv/dt

Solving for v(t) I get: ln[m/([alpha]*b*t)] + vo

b)

When the ship stops.

v(t) = 0 = ln[m/([alpha]*b*t)] + vo
t = 1/(b*[alpha])*m*e^(vo*[alpha])

c)

Find x(t):

v(t) = dx/dt = ln[m/([alpha]*b*t)] + vo

Solving for x(t) = t/[alpha]*{ln(m/([alpha]*b*t) + vo*t + 1} + xo

My teacher said this problem was difficult. However it seems very straight forward to me, unless I'm doing something completely wrong.

Thanks

Tom Mattson
Staff Emeritus
Gold Member
Originally posted by frankR
For a)

Finding v(t)

F = -b*e^([alpha]*v) = m dv/dt

Solving for v(t) I get: ln[m/([alpha]*b*t)] + vo

For starters, you have a glitch here. At some point you would have arrived at the step:

exp(-&alpha;v)=(&alpha;bt/m)+e(-&alpha;v0)

and from there you have to take the natural log to get to v(t). You made a mistake when you did that, namely you employed the step:

ln(a+b)=ln(a)+ln(b),

which is not valid.

I'm not sure what you mean. I have to go to class right now, no matter.

I redid the problem and came to the same answer. I must of made the same mistake.

Sometimes a repeatable result doesn't make you right, it just means you made the same mistake twice.

Tom Mattson
Staff Emeritus
Gold Member
Originally posted by frankR
I'm not sure what you mean. I have to go to class right now, no matter.

OK, what I'm talking about is this:

exp(-&alpha;v)=&alpha;bt/m+exp(-&alpha;v0)

-&alpha;v=ln(&alpha;bt/m)-&alpha;v0.

-&alpha;v=ln(&alpha;bt/m+exp(-&alpha;v0)).

HallsofIvy
Homework Helper
Perhaps the previous comments will make more sense if you can answer this: What value of t are you using for the "initial" time?

If v(t)= ln[m/([alpha]*b*t)] + vo then v(0) is not defined.

What value of t gives v(t)= 0?

You may have solved the d.e. m dv/dt= -be[alpha]v

to get v(t)= ln(m/([alpha]*b*t) and then simply added vo: you can't do that.

mdv/dt= F= -be^([alpha]v) gives e^(-[alpha]v)dv= (-b/m)dt which integrates to (-1/[alpha])e^(-[alpha]v)= (-b/m)t+ C
or e^(-[alpha]v)= (b[alpha]/m) t+ C. (this C is -[alpha] times the old C but since they are both "undetermined constants" I won't make the distinction.)

We can determine C at this point. When t= 0, v= vo so
e^(-[alpha]vo)= C. Our equation is now
e^(-[alpha]v)= (b[alpha]/m)t+ e^(-[alpha]vo)

Taking logarithms of both sides we get

-[alpha]v= ln((b[alpha]/m)t+ e^(-[alpha]vo))

so v= -(1/[alpha])ln((b[alpha]/m)t+ e^(-[alpha]vo))

To determine when the boat will come to a stop, let v= 0:
-(1/[alpha])ln((b[alpha]/m)t+ e^(-[alpha]vo))=0
so (b[alpha]/m)t+ e^(-[alpha]vo)= 1

t= (1- e^([alpha]vo)(m/b[alpha])

v= dx/dt= -(1/[alpha])ln((b[alpha]/m)t+ e^(-[alpha]vo))

To integrate this, let u= (b[alpha]/m)t+ e^(-[alpha]vo))
then du= (b[alpha]/m)dt or dt= m/(b[alpha])du so
dx= -(m/b[alpha]^2)ln(u)du

Of course, the integral of ln(u) is u ln(u)- u so

x(t)= -(m/b[alpha]^2)(u ln(u)- u)+ C
= -(m/b[alpha]^2)(b[alpha]/m)t+ e^(-[alpha]vo))ln((b[alpha]/m)t+ e^(-[alpha]vo))- (b[alpha]/m)t+ e^(-[alpha]vo))))

Tom Mattson
Staff Emeritus
Gold Member
HallsOfIvy,

Apparently, you have chosen not to receive Private Messages, so I have to post this here.

I appreciate all the time you spend here in the Homework forum. I would only ask you to stop posting complete solutions to the problems. I want the students to show their work and where they get stuck, then we can help through the rough spots. If we just give the solution away, then this will become a place where students run to when they don't feel like thinking, and that's not what I want.

Could you please edit the above solution? Then I will delete this post.

Thanks,

Okay.

So:

ln(e^([alpha](v - vo))) != [alpha](v - vo)

ln(e^([alpha](v - vo))) = e^[alpha]v/e^[alpha]vo ?

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