How to Find the Velocity of a Gliding Object After Time t?

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SUMMARY

The discussion focuses on calculating the velocity of a gliding object after a time t, starting with an initial thrust-velocity V. The key equations involve the lift force proportional to the angle of attack and velocity squared, leading to the differential equation: dv/dt = -g/v + a*v. The integration process is crucial, with participants discussing the substitution method to simplify the integral and derive a workable equation for velocity as a function of time. The final goal is to express the velocity after time t in a format similar to the kinematic equation Vf = Vi + at.

PREREQUISITES
  • Understanding of basic physics concepts such as gravity and lift forces.
  • Familiarity with differential equations and integration techniques.
  • Knowledge of kinematic equations and their applications.
  • Proficiency in mathematical notation, particularly in calculus.
NEXT STEPS
  • Study the method of substitution in integral calculus for solving differential equations.
  • Learn about the relationship between lift, angle of attack, and velocity in aerodynamics.
  • Explore the derivation of kinematic equations from differential equations.
  • Practice solving similar physics problems involving forces and motion to reinforce understanding.
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Students studying physics, particularly those focusing on mechanics and aerodynamics, as well as educators seeking to clarify concepts related to motion and forces.

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Homework Statement



Finding the velocity of a gliding object with an initial thrust-velocity of V after time t.
Constants are:-
g: gravity
a: constant for angle of attack, lift coefficient, air density and wing-area

Homework Equations



[tex]L \propto a*V^2[/tex]
acceleration = v (dv/dt)
V is the airspeed of the object

The Attempt at a Solution


I know lift is a force and has an acceleration. So thinking of Lift in terms of velocity-dependent acceleration and only focusing on Vy

acceleration = -g + a*v^2
[tex]v (dv/dt) = -g + a*v^2[/tex]
[tex]dv/dt = -g/v + a*v[/tex]
[tex]dv/((-g/v) + a*v) = dt[/tex]

[tex]\int_{v_0}^v dv/(-g/v + a*v) = \int_{0}^t dt[/tex]

I haven't done integration in a long time.. also I am confused about the notation
Heres the equation I get
[tex]\Big|_{v_0}^v (g*ln((a*v^2)-g) + a*v^2)/(a*v^2) = t[/tex]
After this.. I am unsure on how to proceed further
 
Last edited:
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Try writing your integral as
[tex]\int_{v_0}^v \frac{v}{-g+av^2}dv[/tex]
Then substitute in u=-g+av^2
 
Tomsk said:
Try writing your integral as
[tex]\int_{v_0}^v \frac{v}{-g+av^2}dv[/tex]
Then substitute in u=-g+av^2

thanks for the reply.. after integrating that i get [tex]\Big|_{v_0}^v \frac{ln(a*v^2+g)} {(2*a)} = t[/tex] ..which looks a lot more workable, but is the integration correct?
and how do i proceed from here for coming up with a v = v0(t) equation? (i.e. a "Vf= Vi+at" -type equation where given an initial velocity I can come up with a velocity after time t)
Im not asking about the actual math involved.. I am just unsure as to what this:- " [tex]\Big|_{v_0}^v[/tex] " notation means, since it will determine the initial equation i need to simplify
I tried the following way but..

1) [tex]\Big|_{v_0}^v \frac{ln(a*v^2+g)} {(2*a)} = t[/tex]

2) [tex](v-v0) ( \frac{ln(a*v^2+g)} {(2*a)}) = t[/tex]

3) [tex](v-v0) = \frac{t*2*a} {ln(a*v^2+g)}[/tex]

4) [tex]v = v0 + \frac{t*2*a} {ln(a*v^2+g)}[/tex]

I get stuck again since I have an unknown [tex]v^2[/tex] in the fraction on the right side.. should i try another method from step 2 onwards or is the step wrong to begin with?
 
Last edited:

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