Question

The rear wheels of a sports car have a radius of \(48.95 \mathrm{~cm}\). The sports car accelerates from rest with constant acceleration for \(3.997 \mathrm{~s}\). The angular acceleration of the rear wheels is \(14.99 \mathrm{~s}^{-2} .\) What is the final linear speed of the car? Assume that the wheels roll without slipping.

Short answer

Answer: The final linear speed of the sports car is approximately 29.36 m/s.

Step-by-step solution

Calculate the final angular speed

To get the final angular speed, we use the formula ω = αt, where α is the angular acceleration and t is the time. Plugging in the given values, we get: ω = (14.99 s^{-2})(3.997 s) Calculating the value, we get: ω = 59.94 s^{-1}

Calculate the final linear speed

Now we can use the formula for linear speed: v = Rω, where R is the radius of the wheel and ω is the final angular speed. Plugging in the given value for the radius and the calculated value for ω, we have: v = (48.95 cm)(59.94 s^{-1}) Now we'll convert the radius from centimeters to meters (100 cm = 1 m): v = (0.4895 m)(59.94 s^{-1}) Multiplying the values, we get: v ≈ 29.36 m/s So the final linear speed of the sports car is approximately \(29.36 \mathrm{~m/s}\).