Un objeto tiene una velocidad de vi=3i-4j m/s, luego duplica su velocidad en 12 segundos, calcula la magnitud de la distancia que recorre en metros.

Explanation:

This is an exercise in kinematics, the speed they give is in two dimensions, let's work on each component

X axis

initial velocity v₀ₓ = 3 m / s in a time of t = 12 s, the velocity is doubled, the final velocity is vₓ = 6 m / s

acceleration is

           vₓ = v₀ₓ + aₓ t

           aₓ =

           aₓ = 6 - 3/12

           aₓ = 0.25 m / s²

the distance traveled is

           vₓ² = v₀ₓ² + 2 aₓx x

           x = vx² - vox² / 2a

           x = 6² - 3² / 2 0.25

           x = 54 m

 Y axis

we look for acceleration

          v_y =     v_{oy} + a_y t

          a_y =

          a_y =

          ay = 0.3333 m / s²

the distance is

          v_y² = v_{oy}² + 2 a⁷y

          y = vy² - voy² / 2 0.25

          y = 8² - 4² / 2 0/3333

          y = 72 m

the distance traveled is

           r = (54 i + 72j) m / s

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