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Mechanics11/Seite4/Loesung.1.9

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1.9 Rückwärtsgang

a)v=\frac {\Delta {x}} {\Delta {t}} 

 v=\frac {-168m} {12,5s}
 v=-13.4\frac {m} {s}

b)x(t) = x0 + v * Δt 

 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

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