Question

El vector de posicion de una pelota de basket que se ha lanzado a la canasta viene dado por r(t)=3t i + (6t +5t^2) en unidades S.I . Al calcular el vector desplazamiento entre los instantes T=0.50s y T=1.0s se ontiene

Answer 1

r(t)=3t i + (6t +5t^2)j

r(0.5)=3(0.5) i + (6(0.5) +5(0.5)^2)j

r(0.5)=1.5 i + (6(0.5) +5(0.5)^2)j

r(0.5)=1.5 i + 4.25j

r(1.0)=3(1.0) i + (6(1.0) +5(1.0)^2)j

r(1.0)=3i + 11j

Δr(0.5-1.01) = r(1.0) - r(0.5) = (3i + 11j )- (1.5 i + 4.25j) = 1.5i + 6.75j

Answer 2

(t)=3t i + (6t +5t^2)j

r(0.5)=3(0.5) i + (6(0.5) +5(0.5)^2)j

r(0.5)=1.5 i + (6(0.5) +5(0.5)^2)j

r(0.5)=1.5 i + 4.25j

r(1.0)=3(1.0) i + (6(1.0) +5(1.0)^2)j

r(1.0)=3i + 11j

Δr(0.5-1.01) = r(1.0) - r(0.5) = (3i + 11j )- (1.5 i + 4.25j) = 1.5i + 6.75j

Espero que te sirva