Also dann so?

Text erkannt:
1. geg: v0=9,3 m/s
ωr(t)v(t)∣v(1)∣=7,451=v0⋅t(cos(ωt)sin(ωt))=r˙=(−v(ωtsin(ωt)−cos(ωt))v(sin(ωt)+cotcos(ωt)))=(−9,3⋅(7,4⋅sin(7,4)−cos(7,4)))2+(9,3(sin(7,4)+7,4cos(7,4)))2=69,45mrs
2. a∣a(1)∣=v˙=(−vω(2sin(ωt)+ωtcos(ωt))−vω(ωtsin(ωt)−2cos(ωt)))=(−9,3⋅7,4⋅(2sin(7,4)+7,4⋅cos(7,4)))2+(−9,3⋅7,4(7,4⋅sin(7,4)−2cos(7,4)))2=527,54mas2