User:Alex Bodev

hello my name is alexander and i visit the gisela gymnasium.

WE LOVE YOU SCHATZI *knuuuutsch*

=Sandkasten=It is essential: Start at 168m, x= 0 after 12,5s

Lösung von Aufgabe 1.9.(Rückwärtsgang)
a)$$v=\frac {s}{t}$$ $$=\frac {168m}{12,5s}=-13,44 m/s $$

b)$$=\frac x=168m-13.44m/s173s$$

Schön, dass Sie an der Lösung dieser Aufgabe gearbeitet haben!--White Eagle 12:08, 22 October 2007 (CEST)



Gute Arbeit, weiter so!--White Eagle 11:05, 13 November 2007 (CET)

Aufgabe 3.8
geg: v= 70km/h t= 0.20 s

ges: Verzögerung

a.

Gegeben: $$70\frac{km}{h}=19,4\frac {m}{s} t=0,2s$$

Lösung: $$a=\frac {v}{t}=\frac{19,4\frac {m}{s}}{0,2s}=97,0\frac {m}{s^2}$$

b.

Lösung: $$\frac{F_b}{F_g}= \frac{m*a}{m*g}=\frac{97 \frac{m}{s^2}}{9,81 \frac{kg}{N}}= 9,9$$

Antwort: Der Fahrer erfährt das 9,9 fache seiner Gewichtskraft.

Translation
Exercise 1.9: Reverse Gear

Calculate the ideal movement from the picture below of the t-x diagram, which we can imagine to continue as long as we want

b) speed of vehicle a) the coordinate x(1) of the position, which the vehicle has at the time t(1)= 2min53s c) the time t(2), on which it is at the position with the coordinate x(2)= -6,2km is!

Horizontal axis: time in s, Orthographic in m

Exercise 1.10: Catch up!

A lorry drives on a straight road with the constant speed v(1)= 45m/s from the position with the coordinate 47,7km on.

1.7 Hürdenlauf
x Ort

(1) X(10s) = 4m/s x 10s X = 40m

(2) X = 40m +(-2,0 m/s) x 10s X = 20m

(3) X = 20m +(0 m/s) x 5s X = 20m

(4) X = 20m +(1 m/s) x 5s X = 25m

(5) X = 25m +(2 m/s) x 5s X = 35m

(6) X = 35m +(4 m/s) x 10s X = 75m

(7) X = 75m +(0 m/s) x 5s X = 75m

Antwort:Helena legt auf ihrem Pferd, insgesamt 75 meter zurück

in Zusammenarbeit mit User:A.Abel