Question

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.

Answer 1

Answer:

x = 5 m  ;approximate

Explanation:

Kinematic equation for the object

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

$(v_{f})^{2} =(v_{o})^{2} + 2*a*d$

$v_{f}$ : final speed ( m/s)

$v_{o}$ : 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)

$(v_{f})^{2} = (v_{o} )^{2} +2*a*d$

$d = \frac{6.4^{2}- 4.3^{2} }{2*5}$

d= 2.247 m

Calculation of the final position : $x_{f}$

$x_{f} =x_{o} +d$

$x_{f} = 2.7 m + 2.247 m$

$x_{f} =4.947 m$

$x_{f} = 5m$  : approximate