An object moves in one dimensional motion with constant acceleration a = 5 m/s^2. at time t = 0 s, the object is at x0 = 2.7 m and has an initial velocity of v0 = 4.3 m/s. how far will the object move before it achieves a velocity of v = 6.4 m/s?
your answer should be accurate to the nearest 0.1 m.

x = 5 m  ;approximate

Explanation:

Kinematic equation for the object

Because the object moves with uniformly accelerated motion we apply the following equation:

: final speed ( m/s)

: initial speed ( m/s)

a: acceleration: ( m/s²)

d : distance (m)

Initial Conditions:

t₀=0 , x₀= 2.7 m, v₀= 4.3 m/s

x₀= inicial position

Final  Conditions:

a=  5 m/s² , vf= 6.4 m/s

Calculation of the distance traveled by the object in final condition

We apply the formula (1)

d= 2.247 m

Calculation of the final position :

 : approximate

the correct choice is a .

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