The Starship Enterprise (variable acceleration problem)

In summary, the Enterprise initially moves with speed vo, but hits an intergalactic metero shower and experiences a deceleration force given by: F=-b*e^([alpha]*v)
  • #1
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


When the ship stops.

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


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.

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  • #2
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:


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:


which is not valid.
  • #3
I'm not sure what you mean. I have to go to class right now, no matter.:wink:

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.
  • #4
Originally posted by frankR
I'm not sure what you mean. I have to go to class right now, no matter.:wink:

OK, what I'm talking about is this:


does not lead to:


It leads to:

  • #5
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))))
  • #6

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.

  • #7


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

ln(e^([alpha](v - vo))) = e^[alpha]v/e^[alpha]vo ?
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1. What is the variable acceleration problem on the Starship Enterprise?

The variable acceleration problem on the Starship Enterprise is a hypothetical issue that occurs when the ship is traveling at high speeds and needs to decelerate quickly. This creates a strain on the ship's structure and can potentially cause damage or even destruction.

2. How does the Starship Enterprise handle variable acceleration?

The Starship Enterprise has a system called the "inertial dampeners" which help to counter the effects of variable acceleration. These dampeners work by creating a field that absorbs and redistributes the force of acceleration, reducing the strain on the ship's structure.

3. Can the variable acceleration problem be solved with current technology?

As of now, variable acceleration is still a theoretical problem and has not been encountered in real life. Therefore, there is no need for a solution with current technology. However, there are ongoing research and development efforts to address potential issues with high-speed travel in the future.

4. Are there any potential risks associated with the variable acceleration problem?

Yes, the variable acceleration problem can pose a significant risk to the safety and functionality of the Starship Enterprise. In extreme cases, it could lead to structural damage or even the destruction of the ship. That is why it is crucial for the Enterprise to have advanced inertial dampeners to mitigate these risks.

5. How does the variable acceleration problem affect the crew on the Starship Enterprise?

The variable acceleration problem can have a significant impact on the crew of the Starship Enterprise. It can cause discomfort, disorientation, and even injury if the acceleration is too sudden or severe. The crew must be trained and prepared to handle these potential effects when traveling at high speeds.

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