A particle is in uniform circular motion about the origin of an xy coordinate system, moving clockwise with a period of 3.50 s. At one instant, its position vector (from the origin) is r = (8.00 m) ihat - (7.00 m) jhat. At that instant, what is its velocity in unit-vector notation?

To find the velocity of a particle in uniform circular motion, we need to determine its direction and magnitude.

Since the particle is moving in a circle, its velocity vector is tangent to the circle at any point. In other words, it is always perpendicular to the position vector.

Given that the particle is moving clockwise, the velocity vector points in the opposite direction, which is counterclockwise.

To find the magnitude of the velocity, we can use the formula v = 2πr / T, where v is the magnitude of the velocity, r is the radius of the circular path, and T is the period of the motion.

In this case, the radius of the circular path is the magnitude of the position vector, which is given as:
r = (8.00 m) î - (7.00 m) ĵ

The magnitude of the position vector is:
|r| = √(8.00^2 + 7.00^2) = √(64 + 49) = √113 ≈ 10.63 m

Substituting the values into the formula for velocity, we get:
v = (2π * 10.63 m) / 3.50 s ≈ 60.68 m/s

So the magnitude of the velocity is approximately 60.68 m/s.

The unit vector notation for the velocity vector is found by dividing the original vector by its magnitude:
v̂ = (8.00 m / 10.63 m) î - (7.00 m / 10.63 m) ĵ

Simplifying this expression, we get:
v̂ ≈ 0.75 î - 0.66 ĵ

Therefore, the velocity of the particle in unit-vector notation is approximately 0.75 î - 0.66 ĵ.

The magnitude of the velocity is

|V| = 2 pi R f, where f is the rotation frequency and R is the length of the radius vector. The direction is perpendicular to the radius vector, aimed in the clockwise direction.