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$x (t) = x 0 + v t$

$Δ x = v Δ t$

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

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

$x 1 (t) = x 2 (t)$ </math> $x 1 + v 1 t = x 2 + v 2 t$

$x 1 - x 2 = (v 2 - v 1) t$

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

= Aufgabe 1.10 =

b)

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

$x 1 = x 0 + v t$

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

$x 1 = 266.9 k m$

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

$(1)Δ t$

$(2)Δ v$

$(3) v 0$

$(4) t 0$

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:

Time t in s	Speed v in m/s
0	5	10	60	80	100	110	120	130
0	1	2	12	12	12	8	4	0

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!

User:Chris1727

From WikiEducator

Übersetzung

Problem 2.3: Accelerationtest

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