v^2

Un automobilista sta guidando a una velocità costante di 80 km/h quando vede un ostacolo sulla strada a 50 m. I freni gli consentono di sviluppare una decelerazione  a = - 6 m/s^2. Riuscirà il guidatore a fermarsi prima dell'ostacolo?  (s^2 significa s al quadrato)

Il moto é rettilineo uniformemente decelerato, ossia con accelerazione costante negativa. Diciamo S lo spazio di frenata, conosciamo accelerazione, velocità iniziale e velocità finale, che deve essere zero. Utilizziamo l'equazione:

vfin^2 = viniz^2 + 2aS

Trasformiamo la velocità in m/s: viniz = 80 km/h = 22,222 m/s.

L’equazione precedente diventa: 0 = (22,222 m/s)^2 + 2⋅(-6 m/s^2)⋅S

Quindi: S = 41,2 m

L'ostacolo è a 50 metri e quindi l'automobilista si ferma prima.

Cinematica con dinamica

Un oggetto viene lanciato su una rampa inclinata di 45° con una velocità iniziale di 30 m/s. Dopo quanto tempo si ferma ? A che altezza dal suolo arriva?

Sia O l'origine. Il riferimento sia paralleo al piano inclinato verso l'alto. La direzione di moto è  positiva quando è erso l’alto. La velocità iniziale risulta positiva e l’accelerazione di gravità (diretta verso il basso)
negativa.  Valuatiamo la componente dell’accelerazione lungo la direzione di moto, che risulta essere: a = -g sen α = - 6,94 m/s (con g indico il modulo dell'acc. di gravità, pari a 9.8m/s^2).
Per calcolare il tempo che occorre all’oggetto per fermarsi devo ricordarmi che la velocità finale, in
questo caso, é nulla, e poi usare la definizione di accelerazione:

a = (vfin - v iniz)/ t

Nel nostro caso: -g sen α = - 6,94  m/s^2= (vfin - v iniz)/ t= (0 -  30 m/s) / t

Quindi:

t = 4,32 s

Per calcolare lo spazio S percorso sulla rampa, ho a disposizione due espressioni:

S = 1/2 a t^2 + viniz t

Si ottiene:

S = 1/2 (- 6,94) (4,32)^2 + 30 ⋅ 4,32 =  64,8 m

La quota H raggiunta sul livello del suolo la si ricava come:

H = S sen α = 64,8 m ⋅ sen 45° = 45,8 m

Asta

Esercizio sulla Statica dei Corpi Rigidi,  Prof. T.Papa.

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