am nevoie de acest exercițiu cât se poate de repede! multumesc!
Fizica, cls 9
Răspunsuri la întrebare
Răspuns:
a) τ=59,7s b) t3=5,53s
Explicație:
Notez:
a-acceleratia trenului
L-lungimea vagonului
d0-distanta parcursa de tren pana la ajungerea pasagerului pe peron in fata vagonului
a)
Folosind formula generala pentru miscarea uniform accelerata
X-X0=V0·t+a·t²/2 =>
d0=a·τ²/2
d0+L=a·(τ+t1)²/2 => L=a·(τ+t1)²/2-d0 => L=a·(τ+t1)²/2-a·τ²/2 => L/a=(τ+t1)²/2-τ²/2=τ²/2+τ·t1+t1²/2-τ²/2=τ·t1+t1²/2
d0+2·L=a·(τ+t1+t2)²/2 => 2·L=a·(τ+t1+t2)²/2-d0 => 2·L=a·(τ+t1+t2)²/2-a·τ²/2 => 2·L/a=(τ+t1+t2)²/2-τ²/2=τ²/2+τ·(t1+t2)+(t1+t2)²/2-τ²/2=τ·(t1+t2)+(t1+t2)²/2
Dar L/a=τ·t1+t1²/2 => 2·L/a=2·τ·t1+t1² =>
2·τ·t1+t1²=τ·(t1+t2)+(t1+t2)²/2 => 2·τ·t1-τ·(t1+t2)=(t1+t2)²/2-t1² =>
τ·t1-τ·t2=t1²+t1·t2+t2²/2-t1² => τ·(t1-t2)=t1²/2+t1·t2+t2²/2-t1²=t1·t2+t2²/2-t1²/2=t1·t2-(t1-t2)·(t1+t2)/2 => τ=t1·t2/(t1-t2)-(t1+t2)/2 =>
τ=6,6·6/(6,6-6)-(6,6+6)/2=6,6·6/0,6-12,6/2=66-6,3=59,7s
b)
d0+L=a·(τ+t1)²/2
d0+3·L=a·(τ+t1+t2+t3)²/2
=> (d0+3·L)-(d0+L)=a·(τ+t1+t2+t3)²/2-a·(τ+t1)²/2
=>3·L-L=a·[(τ+t1+t2+t3)²-(τ+t1)² ]/2 =>
2·L/a=[(τ+t1+t2+t3)²-(τ+t1)² ]/2
Dar 2·L/a=2·τ·t1+t1² => 2·τ·t1+t1²=[(τ+t1+t2+t3)²-(τ+t1)² ]/2
=> 4·τ·t1+2·t1²=(τ+t1+t2+t3)²-(τ+t1)²
=> 4·τ·t1+2·t1²+(τ+t1)²=(τ+t1+t2+t3)²
=> (τ+t1+t2+t3)²=4·τ·t1+2·t1²+τ²+t1²+2·τ·t1
=> (τ+t1+t2+t3)²=6·τ·t1+3·t1²+τ²
=> t3=√(6·τ·t1+3·t1²+τ²)-(τ+t1+t2)
=> t3=√(6·59,7·6,6+3·6,6²+59,7²)-(59,7+6,6+6)
t3=√(2364,12+130,68+3564,09)-(72,3)=√6058,89-72,3≅77,83-72,3=5,53s