2

I am reading a report titled ``Definition of safe separation criteria for external store and pilot escape capsules'' in which the author uses following expressions.

From Newton's second law: $$ z \left( t\right) = \frac{1}{m} \int \int\limits_{0}^{t}F \left( \tau \right)d\tau dt$$

From Taylor Series,

$$ F \left(\tau \right) = F \left( 0 \right) + \sum\limits_{n=1}^{\infty} \frac{d^{n}F}{d\tau^n} \Bigg\rvert_{\tau = 0} \frac{\tau^{n}}{n!}$$

Substituting the Taylor series expansion in first equation, we get

$$ z \left( t\right) = \frac{1}{m} \left[ \frac{F \left( 0 \right) t^{2}}{2!} + \sum\limits_{n=1}^{\infty} \frac{d^{n}F}{d\tau^n} \Bigg\rvert_{\tau = 0} \frac{\tau^{n+2}}{\left(n + 2 \right)!}\right] + w \left( 0 \right) t + z \left( 0 \right)$$

Questions:

  1. I am familiar with differntial equation form of Newton's Second law, i.e., $F = m \frac{d^{2}z}{dt^{2}}$ but I don't know how the author obtained integral form, especially the limits of integration.
  2. Since the integral with respect to $\tau$ is a definite integral, why the author has added constant of integration when integrating with respect to $\tau$.
Ali Baig
  • 12
  • 10

2 Answers2

4

Review of Terms

w(t) -> v(t) -> velocity at time t

z(t) -> x(t) -> position at time t

w(0) -> initial velocity


Derivation

We can compute $w(t)$ from force, since $\frac{dw}{dt} = F(t) / m$, so $$w(t) - w(0) = \int_{0}^{t}\left(\frac{1}{m}F(\tau)d\tau\right)$$

Which means $$w(t) = w(0) + \int_{0}^{t}\left(\frac{1}{m}F(\tau)d\tau\right)$$

We can compute $z(t)$ from velocity: $$z(t) - z(0) = \int_{0}^{t} \left( w(\tau) d\tau \right)$$

Where we substitute the above integral expression for $w(t)$: $$z(t) - z(0) = \int_{0}^{t}\left( w(0) + \int_{0}^{t}\frac{1}{m}F(\tau) d\tau\right)d\tau$$

Since $w(0)$ is constant, we can compute its integral in closed form: $w(0) * t$.

$$z(t) = z(0) + w(0)t + \int_{0}^{t}\left(\int_{0}^{t}\frac{1}{m}F(\tau) d\tau\right)d\tau$$

And you can use that Taylor expansion to write an expression for this integral.


Conclusion

  1. The integral form is the double-integral of the differential equation.

  2. The integration constants you are seeing come from the fundamental theorem of calculus — they’re the initial conditions. Recall the classic kinematic equation: $z(t) = z(0) + w_0t + \frac{a_0t^2}{2}$ for the same expression, where acceleration is known to be constant. $w_0$ is $w(0)$. The body’s initial velocity applies a linear offset to its motion

Jacob Panikulam
  • 156
  • 3
2

I have never seen such an expression of the Newton second law. From the first equation you provide, it seems that there are two different time variables, $t$ and $\tau$, which is obviously not the case. From what I get this notation is just a translation of a double integration in time. I agree with you that the limits of the integration are quite confusing. To my mind both integral limits should be 0 and t, and I would rewrite the whole equation this way: $$ z \left( t\right) = \frac{1}{m} \int\limits_{0}^{t} (\int\limits_{0}^{t}F \left( \tau \right)d\tau) d\tau $$ As for your second question, the constants of integration just reflect the physical initial conditions. $w(0)$ is the velocity and $z(0)$ the position at time $t=0$, I do not think it has something to do with the integral being definite.

ClariB
  • 174
  • 5