Physics Asked by manuel459 on February 10, 2021
Let’s say I want to calculate the height of a thrown ball: Let $x”(t)=-g$ and $x(0)=x(T)=0$ and $x'(0)=v_0$.
One could then integrate 2 times and it is done. My professor told me to write it this way, and I just don’t get it: Could somebody help me understand it?:
$$x(t)=int_0^tleft(int_0^s x”(r) mathrm{d}r right)mathrm{d}s=-int_0^t(gs+C_1)mathrm{d}s=-frac{1}{2}gs^2-C_1t-C_2$$
Why from $0$ to $s$? And why are there integration constants?
The choice of integration variables ($r$ and $s$) is just to distinguish which variables are currently being integrated. The choice of letter is arbitrary since $$int_0^1 f(x)dx = int_0^1 f(y)dy.$$ Choosing different letters for different integrals keeps them separated into different operations. The variable $r$ denotes the time when the ball has a certain acceleration. The variable $s$ denotes the time when the ball has a certain velocity. And $t$ denotes the time when the ball has a certain position. You could have also chosen $t_a$ for the acceleration time instead of $r$ and $t_v$ for the velocity instead of $s$ and everything would have worked out the same.
Your result has integration constants because you didn't use your initial conditions after each integral.
Answered by Mark H on February 10, 2021
Put it in this way for easy to grab each step:
$$ tag{1} x(t) = x(0) + int_0^t v(tau) dtau. $$ and $$ tag{2} v(tau) = v(0) + int_0^tau a(xi) dxi. $$
These are basic kinematic relations. Substitute Eq.(2) into Eq.(1):
$$ tag{3} x(t) = x(0) + int_0^t left( v(0) + int_0^tau a(xi) dxi right) dtau. $$
This is the form for double integrrals. Your post expression is not quite right. The integral constant comes out very strangely (because you wrote a definite integal, there should have no integral constants.).
or, you may write an indefinite integrals, then leave the integral constants to be determined by initial conditions.
$$ tag{3} x(t) =int^t dtau left( int^tau a(xi) dxi right). $$
Answered by ytlu on February 10, 2021
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