A race car driver is driving his car at a constant speed of 45.0 m/s on a circular track with a radius of 285 m. (a) What is the angular speed (in rad/s) of the car? rad/s (b) What are the magnitude (in m/s2) and direction of the car's acceleration? magnitude m/s2 direction

A) Angular velocity = 0.1579 rad/s

B) Acceleration = 7.11 m/s² in a direction towards the centre of the circle motion

Explanation:

In circular motion,

linear speed = angular speed x radius of the rotation

Thus,

v = ωr

Where;

v = linear speed (m/s)

ω = angular speed (radians/s)

r = radius of the rotation (m)

We are looking for angular speed, thus; making ω the subject, we have;

ω = v/r

We are given that v = 45 m/s and r = 285m

Thus,

ω = 45/285 = 0.1579 rad/s

B) The formula for Centripetal acceleration is given as;

a_c is given by the expression

a_c = rω² = v²/r

where;

v is linear velocity of the object,

ω is its angular velocity and

r is the radius of the circle in which object moves.

Thus,

a_c = v²/r = 45²/285 = 7.11 m/s²

The centripetal acceleration affects the direction of the car. So in this case the direction of the car will be towards the centre of the circle motion.

(a) 0.158 rad/s.

(b) 7.1 m/s² toward the center of the circular track.

Explanation:

(a)

Using,

v = ωr ... Equation 1

Where v = speed of the car, r = radius of the circular track. ω = angular speed.

make ω the subject of the equation

ω = v/r ... Equation 2

Given: v = 45 m/s, r = 285 m.

substitute into equation 2

ω = 45/285

ω = 0.158 rad/s

Hence the magnitude of the angular speed of the car = 0.158 rad/s.

(b)

Using

a = v²/r ... Equation 3

Where a = acceleration of the car

Given: v = 45 m/s, r = 285 m

Substitute into equation 3

a = 45²/285

a = 2025/285

a = 7.1 m/s² toward the center of the circular track.