- #1
macarroni
- 1
- 0
I don't know how to solve this really easy integral.
suppose: [tex] a=2 m/s^2 [/tex]
[tex] a = dv / dt [/tex]
[tex] \int^b_a a \ dt = \int^b_a dv [/tex]
[tex] \int^b_a 2 \ dt = v_{b} - v_{a} [/tex]
[tex] 2t_{b} - 2t_{a} = v_{b} - v_{a} [/tex]
[tex] v_{b} = v_{a} + 2t_{b} - 2t_{a} [/tex]
I hope everything is ok up to this point. Then I want to know the position: r, and I need to use the same b and a that I have already use before, but...
[tex] v = dr / dt [/tex]
[tex] \int^b_a v \ dt = \int^b_a dr [/tex]
[tex] \int^b_a ( v_{a} + 2t_{b} - 2t_{a} ) \ dt = r_{b} - r_{a} [/tex]
[tex] v_{a}t_{b} - v_{a}t_{b} ... \ ? = r_{b} - r_{a} [/tex]
suppose: [tex] a=2 m/s^2 [/tex]
[tex] a = dv / dt [/tex]
[tex] \int^b_a a \ dt = \int^b_a dv [/tex]
[tex] \int^b_a 2 \ dt = v_{b} - v_{a} [/tex]
[tex] 2t_{b} - 2t_{a} = v_{b} - v_{a} [/tex]
[tex] v_{b} = v_{a} + 2t_{b} - 2t_{a} [/tex]
I hope everything is ok up to this point. Then I want to know the position: r, and I need to use the same b and a that I have already use before, but...
[tex] v = dr / dt [/tex]
[tex] \int^b_a v \ dt = \int^b_a dr [/tex]
[tex] \int^b_a ( v_{a} + 2t_{b} - 2t_{a} ) \ dt = r_{b} - r_{a} [/tex]
[tex] v_{a}t_{b} - v_{a}t_{b} ... \ ? = r_{b} - r_{a} [/tex]