User:Chris1727

=Sandbox=

Text

Kursiv

$$x(t)=x_0 + vt$$

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

$$v = \frac{x}{t}$$

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

$$x_1(t) = x_2(t)$$ $$x_1 + v_1 t = x_2 + v_2 t$$

$$x_1 - x_2 = (v_2 - v_1) t$$

$$t = \frac{x_1 - x_2}{v_2 - v_1} = \frac{173 m - 25 m}{28 km /h - 11 km/h}$$

= Aufgabe 1.10 =

b)

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

$$x_1 = x_0 + v t$$

$$x_1 = 47.7km + 45 \frac{m}{s} 4870s $$

$$x_1 = 266.9km$$

Geschwindigkeit PKW

$$v_2 = \frac{\Delta x }{\Delta t } = \frac{266.9km - 15.2km}{4800s} = 52.4 \frac{m}{s}$$

c)

aus b) folgt:

266.9km - 47.7km = 219.2km = Strecke des LKW

266.9km - 15.2km = 251.7km = Strecke des PKW



=Übersetzung=

It follows in this case:

$$a =\frac{v}{t}$$

(Movement with constant acceleration from that rests)

If the venture has already at the beginning of the movement a beginning speed $$v_0$$ so the functional equation is

$$v(t) = v_0 + a t$$

The graph is a postponed origin-straight



$$(1)\Delta t$$

$$(2)\Delta v$$

$$(3)v_0$$

$$(4) t_0$$

It tourns out for a>0 a movement with acceleration,for a = 0 a movement with constant speed and for a<0 a movement with constant delay(falling graph)

!Attention! This time the formula$$ a = \frac{v}{t}$$ is wrong!!

To be used is:

$$a = \frac{\Delta v}{\Delta t}$$

a is also the gradient of the line

Definition: The accleretation a is the gradient of the t-v-graph

The unit of the acceleration is m/s²

=Problem 2.3: Accelerationtest=

A testinstitution analyses the movement of a vehicle.The results are:

a) Draw a t-v-diagramm! (10s=1 cm, 1m/s=1 cm) b) Name the type of movement in the particular timezones! c) Find out the accelerations by using the graph!