Question

A sports car accelerates from rest to a speed of \(29.53 \mathrm{~m} / \mathrm{s}\) with constant acceleration in \(4.047 \mathrm{~s}\). The angular acceleration of the rear wheels is \(17.71 \mathrm{~s}^{-2} .\) What is the radius of the rear wheels? Assume that the wheels roll without slipping.

Short answer

#Answer# The radius of the rear wheels is approximately 0.41 meters.

Step-by-step solution

Identify the given parameters and the target parameter

In this problem, we're given: - Linear acceleration of the car, a = 29.53 m/s from rest - Time taken for acceleration, t = 4.047 s - Angular acceleration of the rear wheels, α = 17.71 rad/s² - The wheels roll without slipping The target parameter is the radius of the rear wheels, r.

Calculate the final linear speed of the car

As the car is accelerating from rest, we can use the equation: v = u + a*t where, v = final speed = 29.53 m/s (given) u = initial speed = 0 (since at rest) a = acceleration = 29.53 m/s² (given) t = time = 4.047 s (given) Plug in the given values and calculate for v: v = 0 + (29.53 m/s²)(4.047 s) v = 29.53 m/s

Calculate the final angular speed of the rear wheels

The final angular speed, ω, can be calculated using the equation: ω = α*t where ω = final angular speed α = angular acceleration = 17.71 rad/s² (given) t = time = 4.047 s (given) Plug in the given values and calculate for ω: ω = (17.71 rad/s²)(4.047 s) ω = 71.73 rad/s

Use the rolling without slipping property to find the radius

If the wheels roll without slipping, it means there is a relationship between the linear speed of the car and the angular speed of the rear wheels: v = r*ω where v = linear speed = 29.53 m/s r = radius of the rear wheels ω = angular speed = 71.73 rad/s Now, we can solve for r: r = v/ω r = (29.53 m/s)/(71.73 rad/s) r ≈ 0.41 m The radius of the rear wheels is approximately 0.41 meters.