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

The object will move approximately 10.1 m before it achieves a velocity of 6.4 m/s with a constant acceleration of 5 m/s^2.

Step-by-step explanation:

To find the distance the object will move before it achieves a velocity of 6.4 m/s, we can use the equation v^2 = v0^2 + 2ax, where v is the final velocity, v0 is the initial velocity, a is the acceleration, and x is the distance. Rearranging the equation, we get x = (v^2 - v0^2) / (2a). Plugging in the values, we have x = (6.4^2 - 4.3^2) / (2 * 5) = 10.12 m. Therefore, the object will move approximately 10.1 m before it achieves a velocity of 6.4 m/s.

Answer 2

x = 5 m ;approximate

Step-by-step 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 = (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