esercizio, fisica

x(t) = 2,0 t + 1;

y(t) = 5,0 - t;

Per t che va da 0 s a 5 s;

x(0) = 2,0 * 0 + 1 = 1 m;

y(0) = 5,0 - 0 = 5,0 m;  A(1; 5)

x(1) = 2 * 1 + 1 = 3 m;

y(1) = 5,0 - 1 = 4 m; B(3; 4)

x(2) = 2,0 * 2 + 1 = 5 m;

y(2) = 5,0 - 2 = 3 m; C(5; 3)

x(3) = 2,0 * 3 + 1 = 7 m;

y(3) = 5,0 - 3 = 2 m; D(7; 2)

x(4) = 2,0 * 4 + 1 = 9 m;

y(4) = 5,0 - 4 = 1 m; E(9;1)

x(5) = 2,0 * 5 + 1 = 11 m;

y(5) = 5,0 - 5 = 0 m; F(11; 0)

Vettore posizione A = radicequadrata(1^2 + 5^2) = radice(26) m;

Ciao @anna-sa91

x(0) = 2.0*0 + 1 = 1  , x(1) = 2.0*1 + 1 = 3.0 , x(2) = 2.0*2 + 1 = 5.0 , x(3) = 2.0*3 + 1 = 7.0 , x(4) = 2.0*4 + 1 = 9.0 , x(5) = 2.0*5 + 1 = 11.0

y(0) = 5.0 - 0 = 5.0  , y(1) = 5.0 - 1 = 4.0  ,  y(2) = 5.0 - 2 = 3.0  ,  y(3) = 5.0 - 3 = 2.0  ,  y(4) = 5.0 - 4 = 1.0  ,  y(5) = 5.0 - 5 = 0.0

A(1,5) ; B(3,4) ; C(5,3) ; D(7,2) ; E(9,1) ; F(11,0)

OA =sqrt(1²+5²) =~ 5.099

OB = sqrt(3²+4²) = 5

OC = sqrt(5²+3²) = sqrt(34) = ~ 5.8309

OD = sqrt(7²+2²) = sqrt(53) = ~ 7.280

OE = sqrt(9^2 + 1^2) = sqrt(82) = ~ 9.05538

OF = sqrt(11²+0²)  = 11

x(t) = 2*t + 1  ---> x = 2(5 - y) + 1 --->   y = (11 - x)/2

y(t) = 5 - t   ---> t =5-y