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I'm trying to find a way to calculate a constant acceleration given the following:

  • One dimensional motion.
  • The only forces involved are the force ($F_a$) caused by a constant acceleration ($a_c$) and the drag ($F_d$) against it.
  • $a_c$ - Constant acceleration
  • $v_o$ - Initial velocity
  • $v_t$ - Terminal velocity
  • $m$ - Mass of object
  • $\rho$ - Density of fluid
  • $A$ - Projected area of the object
  • $C_d$ - Drag coefficient
  • $t_t$ - Time to reach n% (e.g. 95%) of terminal velocity

  • $v_o = 0\,m/s$
  • $v_t = 300\,m/s$
  • $m = 1000\, kg$
  • $\rho = 1\, kg/m^3$
  • $A = 100\, m^2$
  • $C_d = 0.5$
  • $t_t = 4\, sec$

    Terminal Velocity: $$ v_t = \sqrt{2 m a_c \over \rho A C_d} $$

    Drag Force: $$ F_d = {1\over2}\rho A C_d v^2 $$

    Solving for $a_c$ using the terminal velocity equation is easy enough ($ a_c = { \rho A C_d v_t^2 \over 2 m } = 2250\,m/s^2$), but that doesn't correctly account for the target time ($t_t$).

    How can I find the correct $a_c$ to reach n% of $v_t$ at time $t_t$? I'm assuming $m$, $A$, and/or $C_d$ will need to be adjusted to achieve this, but I'm having trouble figuring out how to bring the target time concept into the mix in the first place.

    Update: Related question: Calculating time to reach certain velocity with drag force, but 1) I'm having trouble fully grasping the math that's going on there. 2) It's solving for the time required to reach a specific velocity, whereas I'm trying to find the necessary constant acceleration to reach a specific velocity at a specific time.

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    closed as off-topic by John Rennie, Kyle Kanos, ACuriousMind, Danu, Gert Mar 26 '16 at 2:22

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    • $\begingroup$ It doesn't work the way you seem to be assuming. You need to set up an equation of motion, depending on the forces that act on the object. That will give you an expression for the acceleration. Integrate it to find an expression for speed. $\endgroup$ – Gert Mar 24 '16 at 1:33
    • $\begingroup$ Not sure what you're trying to do. I looked up the equation for terminal velocity. The "constant acceleration" $a_c$ in your terminal velocity expression is supposed to be the acceleration due to gravity. Don't understand why you are trying to vary that. $\endgroup$ – Samuel Weir Mar 24 '16 at 1:43
    • $\begingroup$ Welcome to Physics.SE! Some research should convince you that an object falling in fluid approaches terminal velocity with varying acceleration. I believe the approach is asymptotic but don't know the exact form. Should your "constant acceleration" be the acceleration due to gravity? $\endgroup$ – rob Mar 24 '16 at 2:06
    • $\begingroup$ Acceleration due to gravity could be an example value for $a_c$, but I deliberately kept it more generalized because it needs to apply to other forms of acceleration (e.g. the gravitational acceleration of a non-earth planet, or the thrust of self-propelled object moving in a straight line). $\endgroup$ – ustor Mar 24 '16 at 2:48
    • $\begingroup$ Somewhat related: Calculating time to reach certain velocity with drag force, but 1) I'm having trouble fully grasping the math that's going on there. 2) It's solving for the time required to reach a specific velocity, whereas I'm trying to find the necessary constant acceleration to reach a specific velocity at a specific time. $\endgroup$ – ustor Mar 24 '16 at 3:14
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    Step One: Writing Equation of Motion

    Step Two: Solving Equation of Motion for Velocity

    Both these steps are shown in the following Mathematica link where the Equation is typed in the upper portion of the link and the functional form of the velocity as a function of time is calculated:

    Equation of Motion and Its Solution

    Known Quantities Plugged In

    Using Initial Value Condition v(0)=0

    V(t) = V_Terminal * Tanh(0.158114*Sqrt(a)*t)

    ,where V_terminal = 6.32456*Sqrt(a) Since taking the limit of velocity for very large time scales would identically be equal to unity.

    Now you can use these last two equations knowing n% of V_terminal obtained after t_t seconds.

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    • $\begingroup$ Excellent! Thank you for taking the time to break this down into manageable steps. $\endgroup$ – ustor Mar 24 '16 at 18:20
    • $\begingroup$ No problem. I was myself curious to see how would velocity function look like with such an interesting drag force model. Not every drag force model would give us such a convergent pattern velocity. $\endgroup$ – Benjamin Mar 24 '16 at 22:41

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