A particle is traveling counterclockwise in a circle of radius

r = 2.85 m.

At some instant in time, the particle is located by the angular coordinate

α = 34.0°,

the total acceleration has a magnitude of

a = 19.0 m/s2

and is directed at an angle

β = 20.0°

with respect to the radial coordinate. Determine the following at this instant. (Express your answer in vector form.)
determine the velocity and acceleration

To determine the velocity and acceleration of the particle at the given instant, we can break them down into the radial and tangential components.

1. Velocity:
The radial component of velocity (Vr) represents the component of velocity parallel to the radius of the circle.

Vr = ω * r

where ω is the angular speed and r is the radius of the circle.

The tangential component of velocity (Vt) represents the component of velocity tangent to the circle.

Vt = r * dθ/dt

where dθ/dt is the rate at which the angular coordinate changes with time.

To get the total velocity V, we can use the Pythagorean theorem:

V = √(Vr^2 + Vt^2)

2. Acceleration:
The radial component of acceleration (ar) represents the component of acceleration parallel to the radius of the circle.

ar = ω^2 * r

The tangential component of acceleration (at) represents the component of acceleration tangent to the circle.

at = r * d^2θ/dt^2

where d^2θ/dt^2 is the second derivative of the angular coordinate with respect to time.

To get the total acceleration a, we can use the Pythagorean theorem:

a = √(ar^2 + at^2)

Now, let's calculate the velocity and acceleration vector components using the given values:

ω = dθ/dt
Given that α = 34.0°, we know that α = dθ/dt.

Vr = ω * r
Vr = (34.0°) * (2.85 m)
Vr = 96.9 m/s

Vt = r * dθ/dt
Vt = (2.85 m) * (34.0°)
Vt = 96.9 m/s

V = √(Vr^2 + Vt^2)
V = √((96.9 m/s)^2 + (96.9 m/s)^2)
V = √(2 * (96.9 m/s)^2)
V = √(2) * (96.9 m/s)
V ≈ 137.2 m/s

ar = ω^2 * r
ar = (34.0°)^2 * (2.85 m)
ar = 3197.4 m/s^2

at = r * d^2θ/dt^2
Given that β = 20.0°, we know that β = d^2θ/dt^2.

at = (2.85 m) * (20.0°)
at = 57.0 m/s^2

a = √(ar^2 + at^2)
a = √((3197.4 m/s^2)^2 + (57.0 m/s^2)^2)
a = √(10226518.82 m^2/s^4 + 3249 m^2/s^4)
a = √(10229767.82 m^2/s^4)
a ≈ 3199.9 m/s^2

Therefore, the velocity vector at this instant is V = 137.2 m/s (counterclockwise), and the acceleration vector is a = 3199.9 m/s^2 at an angle of 20.0° counterclockwise with respect to the radial coordinate.