User:Benno

hello, my name is Benno i'm 16 years old and i'm a student at the gisela- gymansium (secondary school) in munich

=Sandbox=

'''Ein Elch springt hoch ,  Ein Elch springt weit , ' Warum auch nicht er hat doch zeit ''

x(t)=x0 +vt

$$\Delta x$$=v $$\Delta t$$

v=$$\frac x t$$

v=$$\frac {x_1-x_0}{\Delta t}$$

x1(t)=x2(t)

x1-x2=(v2-v1)t

t=$$\$$

Arbeitsauftrag 19. 10 .07
=1.7 Hürdensprung= 1.7a)

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

Schön, dass Sie an der Lösung dieser Aufgabe gearbeitet haben!--White Eagle 12:07, 22 October 2007 (CEST)

=Graph=



Übersetzung
=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 to.

a)	vehicle speed

b)	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

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

Horizontal axis: time in 2, Orthographic axis in m

Lead

Solution

=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 47,7 km on.