A Simple(?) Differential Equation

[blintaro]

This is question 1.4 of Chapter 8 of Mary L. Boas's Mathematical Methods in the Physical Sciences, Edition 2. I'm using it as a substitute for my ordinary differential equations class since my textbook has apparently been lost somewhere in the mail.

Homework Statement

Find the distance which an object moves in time t if it starts from rest and has an acceleration d^2/dt^2 = ge^(-kt). Show that for a small t the result is approximately (x = 1/2(gt^2)) and for very large t, the velocity dx/dt is approximately constant. (This problem corresponds roughly to the motion of a parachutist.)

Homework Equations

The solution manual gives the solution as: x=k^(-1)gt + k^(-2)g(e^(-kt)-1)

The Attempt at a Solution

I integrated both sides of d^2/dt^2 = ge^(-kt) twice with respect to t, ending up with x = k^(-2)ge^(-kt) + v_0(t) + x_0.
However substituting t=0 into the equation gives x = g/(k^2) + x_0, not (1/2)gt^2, so, obviously on the wrong track.

I put the process in an attached image below since it might be easier to read that than my terrible equation formatting. Thanks!

 

[LCKurtz]

This is question 1.4 of Chapter 8 of Mary L. Boas's Mathematical Methods in the Physical Sciences, Edition 2. I'm using it as a substitute for my ordinary differential equations class since my textbook has apparently been lost somewhere in the mail.

Homework Statement

Find the distance which an object moves in time t if it starts from rest and has an acceleration d^2/dt^2 = ge^(-kt). Show that for a small t the result is approximately (x = 1/2(gt^2)) and for very large t, the velocity dx/dt is approximately constant. (This problem corresponds roughly to the motion of a parachutist.)

Homework Equations

The solution manual gives the solution as: x=k^(-1)gt + k^(-2)g(e^(-kt)-1)

The Attempt at a Solution

I integrated both sides of d^2/dt^2 = ge^(-kt) twice with respect to t, ending up with x = k^(-2)ge^(-kt) + v_0(t) + x_0.
However substituting t=0 into the equation gives x = g/(k^2) + x_0, not (1/2)gt^2, so, obviously on the wrong track.

I put the process in an attached image below since it might be easier to read that than my terrible equation formatting. Thanks!

Please post your equations instead of images of them as per forum rules. Your integrations are not correct. For example, if $x''(t) = ge^{-kt}$, then $x'(t) = -\frac g k e^{-kt} + C$. When you put $t=0$ in both sides you do not get $C = v_0$.

 

[STEMucator]

Here are the equations in a latex format:

$\vec a = ge^{-kt}$
$\vec v = - \frac{g}{k} e^{-kt} + c_1$
$\vec r = \frac{g}{k^2} e^{-kt} + c_1t + c_2$

Using the initial condition $r(0) = 0$, you should get something useful.

$- \frac{g}{k^2} = c_2$

The initial condition that $v(0) = 0$ also gives you some more information.

$\frac{g}{k} = c_1$

Then:

$\vec r = \frac{g}{k^2} e^{-kt} + \frac{g}{k} t - \frac{g}{k^2}$

 

[blintaro]

Oh, I see! So that's how to deal with the c's that wind up after integration. Excellent, thank you very much!