A particle starts from the origin at t = 0 with an initial velocity of 8.0 m/s along the positive x axis. If the acceleration is (-3.7 + 2.5 ) m/s2, determine the velocity and position of the particle at the moment it reaches its maximum x coordinate.

To determine the velocity and position of the particle at the moment it reaches its maximum x coordinate, we first need to find the time it takes for the particle to reach that point.

Let's denote the maximum x coordinate as x_max. To find the time at which the particle reaches x_max, we can use the kinematic equation:

x = x_0 + v_0t + (1/2)at^2

In this equation, x is the displacement, x_0 is the initial position, v_0 is the initial velocity, a is the acceleration, and t is the time.

Since the particle starts from the origin, x_0 = 0. We can substitute the known values:

x_max = 0 + (8.0 m/s)(t) + (1/2)(-3.7 m/s^2)(t^2)

Simplifying this equation, we have:

x_max = 8.0t - 1.85t^2

To find the time when the particle reaches x_max, we can differentiate this equation with respect to t and set it equal to zero:

dx_max/dt = 8.0 - 3.7t = 0

Solving this equation, we find:

t = 8.0 / 3.7 ≈ 2.162 seconds

Now that we have the time at which the particle reaches x_max, we can calculate the velocity and position at that moment.

To find the velocity, we can use the equation:

v = v_0 + at

Substituting the known values:

v = 8.0 m/s + (-3.7 m/s^2)(2.162 s)

Calculating this, we have:

v ≈ 8.0 m/s - 7.99874 m/s ≈ 0.00126 m/s

Therefore, the velocity of the particle at the moment it reaches x_max is approximately 0.00126 m/s.

To find the position, we can substitute the time t = 2.162 s into the equation for position:

x = x_0 + v_0t + (1/2)at^2

Substituting the known values:

x = (8.0 m/s)(2.162 s) + (1/2)(-3.7 m/s^2)(2.162 s)^2

Calculating this, we have:

x ≈ 17.296 m - 17.261 m ≈ 0.035 m

Therefore, the position of the particle at the moment it reaches x_max is approximately 0.035 m.

So, the velocity of the particle is approximately 0.00126 m/s and its position is approximately 0.035 m at the moment it reaches its maximum x coordinate.

Never mind! Figured it out!