A particle confined to motion along the x axis moves with constant acceleration from x = 2.0 m to x = 8.0 m during a 2.5-s time interval. The velocity of the particle at x = 8.0 m is 2.8 m/s. What is the acceleration during this time interval?

with constant acceleration from x = 2.0 m to x = 8.0 m during a 5.3-s time interval. The velocity of the particle at x = 8.0 m is 18.1 m/s. What is the acceleration during this time interval?

0.32

To find the acceleration during this time interval, we can use the following kinematic equation:

Δx = v₀t + (1/2)at²

Where:
Δx is the displacement (change in position),
v₀ is the initial velocity,
t is the time, and
a is the acceleration.

Given information:
Initial position (x₀) = 2.0 m
Final position (x) = 8.0 m
Time (t) = 2.5 s
Final velocity (v) = 2.8 m/s

We need to find the acceleration (a).

First, let's rearrange the equation to solve for acceleration (a):

Δx = v₀t + (1/2)at²
Δx - v₀t = (1/2)at²
2(Δx - v₀t) = at²
2Δx - 2v₀t = at²
a = (2Δx - 2v₀t) / t²

Substituting the given values:

a = (2(8.0 m - 2.8 m/s * 2.5 s)) / (2.5 s)²
a = (2(8.0 m - 7.0 m)) / (2.5 s)²
a = (2(1.0 m)) / (2.5 s)²
a = (2.0 m) / (6.25 s²)
a = 0.32 m/s²

Therefore, the acceleration during this time interval is 0.32 m/s².

v = Vo + a t

x = Xo + Vo t + .5 a t^2

at 8 m
2.8 = Vo + 2.5 a
8 = 2 + 2.5 Vo + .5 a (6.25)

so
Vo = 2.8 -2.5 a
8 = 2 + 2.5(2.8-2.5a) + 3.125 a
solve for a