At some instant, a particle traveling in a horizontal circular path of radius 7.50 m has a total acceleration with a magnitude of 16.0 m/s^2 and a constant tangential acceleration of 12.0 m/s^2. Determine the speed of the particle at this instant and (1/8) revolution later.

I found the first velocity but am struggling with the second.

To find the speed of the particle at the instant, we'll use the equations of motion for circular motion. At any given instant, the total acceleration of the particle is the vector sum of the centripetal acceleration and the tangential acceleration.

Given:
Radius of the circular path (r) = 7.50 m
Total acceleration (|a|) = 16.0 m/s^2
Tangential acceleration (at) = 12.0 m/s^2

The centripetal acceleration (ac) is given by the formula:
ac = v^2 / r

The total acceleration (|a|) can be related to the tangential acceleration (at) and the centripetal acceleration (ac) as follows:
|a| = √(ac^2 + at^2)

We can square both sides:
|a|^2 = ac^2 + at^2

Substituting the known values:
16.0^2 = ac^2 + 12.0^2
256.0 = ac^2 + 144.0
ac^2 = 256.0 - 144.0
ac^2 = 112.0

Now, let's solve for the centripetal acceleration (ac):
ac = √112.0
ac ≈ 10.583 m/s^2

To find the speed (v) at the instant, we'll rearrange the equation for centripetal acceleration:
ac = v^2 / r

v^2 = ac * r
v^2 = 10.583 * 7.50
v^2 ≈ 79.3725

Taking the square root:
v ≈ √79.3725
v ≈ 8.908 m/s (rounded to three decimal places)

So, the speed of the particle at the instant is approximately 8.908 m/s.

Now, to find the speed (v') after a (1/8) revolution later, we know that the tangential acceleration (at) is constant, and no centripetal acceleration is involved. Hence, the tangential acceleration will only change the speed of the particle.

Using the formula for uniformly accelerated motion:
v' = v + at * t

Given:
t = (1/8) revolution = (1/8) * 2π radians

Substituting the known values:
v' = 8.908 + 12.0 * (1/8) * 2π

Simplifying:
v' = 8.908 + 6.2832
v' ≈ 15.191 m/s (rounded to three decimal places)

Therefore, the speed of the particle after a (1/8) revolution later is approximately 15.191 m/s.