Átlagsebesség

1,

A labda útja 2+1 = 3 m

A labda elmozdulása 2-1 = 1 m

2,

s = 10 m

t = 20 s

v = `s/t` = `10/20` = 0,5 `m/s`


3,

s = 8 km = 8000 m

v = 40 `m/s`

t = `s/v` = `8000/40` = 200 s = 3 perc 20 másodperc alatt tette meg.

4,

v = 108 `(km)/h` = 30 `m/s`

t = 5 s

s = `v*t` = `30*5` = 150 m-t tett meg.

5,

t = 10 s

s = 150 m

v = `s/t` = `150/10` = 15 `m/s` = `15*3.6` = 54 `(km)/h` a sebessége.

6,

`v_1` = 40 `(km)/h`

`t_1` = `(s/2)/v_1` = `s/80` s

v = `s_ö/t_ö` = `s/(v_1/(s/2)+v_2/(s/2))`

60 = `cancel(s)/(40/(cancel(s)/2)+v_2/(cancel(s)/2))` = `1/(1/80+1/(2v_2))`

`1/60=1/80+1/(2v_2)`

`1/(2v_2)` = `1/60-1/80` = `(4-3)/240` = `1/240`

`2v_2=240`

`v_2` = 120 `(km)/h` a sebessége visszaúton.


7,

v = 180 `(km)/h` = 50 `m/s`

t = 10 s

a = `v/t` = `50/10` = 5 `m/s^2`

s = `a/2*t^2` = `5/2*10^2` = 250 métert tett meg.

8,

t = 4 s

v = `g*t` = `10*4` = 40 `m/s` sebességgel ér a földre.

s = `g/2*t^2` = `10/2*4^2` = 80 métert tett meg.


9,

r = 200 m

`v_("ker")` = 108 `(km)/h` = 30 `m/s`

`a_(cp)` = `v^2/r` = `30^2/200` = 4,5 `m/s^2`

10,

I. `v_1` = `v_(cs)+v_f=14`

II. `v_2` = `v_(cs)-v_f = 6`

I. - II.:

`2*v_f` = 14-6 = 8

`v_f` = 4 `m/s` a folyó sebessége

b,

I. `v_(cs)+4 = 14`

`v_(cs)` = 10 `m/s` a csónak sebessége.

c,

s = 30 km = 30 000 m

t = `t_1+t_2` = `s/v_1+s/v_2` = `30000/14+30000/6` = 7142 s = 1 óra 59 perc 2 másodperc alatt teszi meg az utat oda-vissza.