A certain wheel turns through 90 rev in \(15 \mathrm{~s}\), its angular speed at the end of the period being \(10 \mathrm{rev} / \mathrm{s} .(a)\) What was the angular speed of the wheel at the beginning of the 15 -s interval, assuming constant angular acceleration? \((b)\) How much time had elapsed between the time the wheel was at rest and the beginning of the 15 -s interval?

Use the formula for final speed in uniformly accelerated motion \( v_f = v_i + at \) which becomes \( \omega_f = \omega_i + \alpha t \), where \(\omega_f\) is the final angular speed, \(\omega_i\) is the initial angular speed, \(\alpha\) is the angular acceleration, and t is the time. We are given that \(\omega_f = 10 \)\( \mathrm{rev/s} \), t= 15 s and \(\omega_i\) has to be found. First we need to calculate the total angular distance covered (D) by the wheel from its formula \( D = \omega_i t + 0.5 \alpha t^2 \). We know D = 90 rev, t = 15 s and need to find \(\alpha\). And from calculus we know \( \omega_f = \omega_i + \alpha t \) so \( \alpha = (\omega_f - \omega_i) / t \) and we can isolate this to find \(\alpha = (D - \omega_i t) / 0.5 t^2 \).

Next, generate an equation system based on the two equations for \(\alpha\) and solve for \(\omega_i\). The equation system would look like: \( \omega_i = 10 - 15 \alpha \) and \( \alpha = (90 - \omega_i * 15) / (0.5 * 15^2) \). After substitution and simplification, one finds that the initial angular speed, \(\omega_i\), was 4 revolution per second.

For this step use the formula for final speed, \( \omega_f = \omega_i + \alpha t \). But when the wheel was at rest, \(\omega_f = 0 \mathrm{rev/s}\). So the formula becomes \(0 = 4 + \alpha t \). Solving this equation yields the time elapsed before the wheel was set into motion. Here, \(\alpha\) is the angular acceleration calculated in the first step. After substitution, you get that the time was approximately 1.6 seconds.