Question

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.

Answer 1

Step-by-step 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 =
`\frac{v_x - v_(ox)} {t}`

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 =
`(v_y - v_(oy) )/(t)`

a_y =
`\frac{8 -4} {12}`

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