How Does Air Resistance Affect Vehicle Acceleration and Deceleration?

AI Thread Summary
Air resistance significantly impacts vehicle acceleration and deceleration, particularly when it is proportional to the square of the vehicle's speed. The discussion revolves around deriving equations to show how a vehicle can reach half its maximum speed in a specific time frame and how it can subsequently come to a stop due to friction and air resistance. The integration of motion equations is explored, with attempts to express acceleration in terms of velocity and time. There is a focus on eliminating constants to relate them to the vehicle's maximum speed. The conversation emphasizes the need for accurate mathematical manipulation to solve the problem effectively.
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Homework Statement



A vehicle of mass m experiences a constant frictional resistance ma and air resistance proportional to the square of its speed. It can exert a constant propelling force mb and attain a maximum speed V Show that, starting from rest, it can attain the speed V/2 in the time

Vln3/2(b-a)​

And that the friction and air resistance alone can then bring it to rest in a further time

(V/(a(b-a))^1/2)tan-1(b-a/4a)^1/2​

2. The attempt at a solution

Using mc as the constant of proportionality for air resistance yields:

x''=b-a-c(x')^2

from there I'm just guessing as to the method but I've tried this:

dx'/dt=b-a-c(x')^2

int(dx')=int((b-a-c(x')^2)dt)
=(b-a)int(dt)-c(int((x')^2dt)
=(b-a)int(dt)-c(int(x'dx))

but I'm not sure how to integrate x' w.r.t. x?

Of course I could be going about it completely the wrong way. The motion is in one dimension so I could maybe import one of the equations of motion, like v^2=u^2+2as? I'm sure I'm missing a trick (or two) somewhere.
 
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Use auxiliary variable v = dx/dt

Then your equation is of the form

\frac{dv}{dt}= A - B v^{2}

from which

\frac{dv}{A-Bv^{2}}=dt

The variables are now separated and you can integrate.
 
integrating:

t=arctan(((-c^1/2)/((b-a)^1/2))v)/(b-a)^1/2(-c^1/2)

rearranging:

v=(((b-a)^1/2)/(-c^1/2))tan((b-a)^1/2)(-c^1/2)t

sorry about the horrible presentation I don't know how to use LaTeX.

The -c^1/2 makes me a little uneasy but I've used the tan form of the integral as it's easier to remember than tanh and judging from the question more likely to yield the result I'm looking for.

OKay so now how do I actually answer the question?
 
I don't know how you got your answer, but I shoved the integral in Mathematica and got

t = \frac{1}{\sqrt{AB}}ArcTanh(\frac{\sqrt{B}}{\sqrt{A}}v)

where A = a - b and B = c. Recheck your work.

To complete the problem, you need to eliminate c in favor of V, i.e. find c in terms of V and the other constants. Hint: What does your diff. eq. look like when the object reaches terminal velocity?
 
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