Mechanics11/Seite4/Loesung.1.9

1.9 Rückwärtsgang

a)v=$$\frac {\Delta {x}} {\Delta {t}}$$ $$v=\frac {-168m} {12,5s}$$  $$v=-13.4\frac {m} {s}$$

b)$$x(t)=x_0+v*\Delta {t}$$ $$x(t)=168m-13.4\frac {m} {s} * 173s$$ $$x(t)=2.16\frac {km} {h}$$

c)$$t=\frac {x(t)-x_0} {v}$$ $$\frac {-6200m-168m} {-13.4\frac {m} {s}}$$ $$=473.8s$$

a) v = $$\frac{168 m}{12,5 s}$$

= -13,4 $$\frac{m}{s}$$

b) x(t) = 168 m - 13,4 $$\frac{m}{s}$$ * 173 s = -2,16 km

c) t = $$\frac{-6200 m -168 m}{-13,4 m /s}$$

= 475 s