How Does Air Resistance Affect Vehicle Acceleration and Deceleration?

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SUMMARY

The discussion focuses on the mathematical modeling of vehicle acceleration and deceleration under the influence of air resistance and friction. A vehicle of mass m experiences a constant frictional resistance represented by ma and air resistance proportional to the square of its speed. The derived equations show that the vehicle can reach half of its maximum speed V in a time of Vln3/2(b-a) and can subsequently come to rest due to friction and air resistance in a time of (V/(a(b-a))^1/2)tan-1(b-a/4a)^1/2. The integration techniques discussed include separating variables and using auxiliary variables to solve the differential equations.

PREREQUISITES
  • Understanding of differential equations and integration techniques
  • Familiarity with vehicle dynamics and forces acting on a vehicle
  • Knowledge of air resistance modeling, particularly quadratic drag
  • Experience with mathematical software such as Mathematica for solving complex integrals
NEXT STEPS
  • Study the derivation of equations of motion under non-linear forces
  • Learn about the application of the ArcTanh function in physics problems
  • Explore the concept of terminal velocity and its implications in vehicle dynamics
  • Investigate numerical methods for solving differential equations in vehicle motion
USEFUL FOR

This discussion is beneficial for physics students, automotive engineers, and anyone interested in the mathematical modeling of vehicle dynamics, particularly in understanding the effects of air resistance and friction on acceleration and deceleration.

Gwilim
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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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