A particle is moving clockwise around a circle with a radius 3.31 m. The magnitude of the total acceleration of the particle at the given instant is 14.3 m/s^2; the angle between the vector of total acceleration and the radius-vector of the particle is 30◦.

Find the centripetal acceleration, tangential acceleration and speed of the particle at the given instant.

a(centr) = a•cosα=14.3•cos30=…

a(tang)=a•sinα=14.3•sin30=…
a(centr)= v²/R,
v=sqrt{a(centr) •R}=...

Thank you!

To solve this problem, we can break down the total acceleration into its components: centripetal acceleration and tangential acceleration.

1. Centripetal Acceleration: This component of acceleration is directed towards the center of the circle and keeps the particle moving in a circular path.

The magnitude of the centripetal acceleration (ac) can be found using the formula:
ac = (v^2) / r

Where v is the velocity of the particle and r is the radius of the circle.

2. Tangential Acceleration: This component of acceleration is directed tangent to the circle and represents any change in the particle's speed.

The magnitude of the tangential acceleration (at) can be found using the formula:
at = a * sinθ

Where a is the magnitude of the total acceleration and θ is the angle between the total acceleration and the radius-vector.

3. Speed of the Particle: The speed of the particle can be found using the formula for centripetal acceleration:
ac = (v^2) / r

By rearranging the equation, we get:
v = sqrt(ac * r)

Now, let's calculate the values:

Given:
Radius (r) = 3.31 m
Magnitude of Total Acceleration (a) = 14.3 m/s^2
Angle (θ) = 30°

1. Centripetal Acceleration:
ac = (v^2) / r

2. Tangential Acceleration:
at = a * sinθ

3. Speed of the Particle:
v = sqrt(ac * r)

By substituting the known values into the equations and performing the calculations, we can find the centripetal acceleration, tangential acceleration, and speed of the particle at the given instant.