91Ó°ÊÓ

The problem provides the angular acceleration function \( \alpha = \sqrt{8t + 1} \). We know that angular acceleration is the derivative of angular velocity, \( \omega \), i.e., \( \alpha = \frac{d\omega}{dt} \). Our goal is to find \( \theta \), the angular displacement, in terms of time \( t \). We have initial conditions that \( \omega = 0 \) and \( \theta = 0 \) when \( t = 0 \).

Integrate the angular acceleration \( \alpha \) with respect to time \( t \) to find angular velocity \( \omega \). \[\omega = \int \alpha \, dt = \int \sqrt{8t + 1} \, dt\]Using the substitution method, let \( u = 8t + 1 \), then \( du = 8 \, dt \), and \( dt = \frac{du}{8} \).Thus, the integral becomes: \[\omega = \int \sqrt{u} \frac{du}{8} = \frac{1}{8} \int u^{1/2} \, du\]Performing the integration, \[\omega = \frac{1}{8} \left( \frac{2}{3} u^{3/2} \right) + C_1 = \frac{2}{24} (8t + 1)^{3/2} + C_1\]Simplify to obtain: \[\omega = \frac{1}{12} (8t + 1)^{3/2} + C_1\] Using the initial condition \( \omega(0) = 0 \), we find:\[0 = \frac{1}{12} (8 \times 0 + 1)^{3/2} + C_1\]\[0 = \frac{1}{12} \cdot 1 + C_1\] \[C_1 = -\frac{1}{12}\] Thus, \[\omega = \frac{1}{12} (8t + 1)^{3/2} - \frac{1}{12}\]

Since angular displacement \( \theta \) is the integral of angular velocity with respect to time, integrate \( \omega \) to find \( \theta \). \[\theta = \int \omega \, dt = \int \left( \frac{1}{12} (8t + 1)^{3/2} - \frac{1}{12} \right) dt\]Breaking it down, \[\theta = \frac{1}{12} \int (8t + 1)^{3/2} \, dt - \int \frac{1}{12} \, dt\]For the first integral, use substitution \( u = 8t + 1 \); hence, the substitution provides the same previous integral setup leading to:\[\theta = \frac{1}{12} \cdot \frac{2}{15} (8t + 1)^{5/2} - \frac{1}{12} t + C_2\]Simplify to get: \[\theta = \frac{1}{90} (8t + 1)^{5/2} - \frac{1}{12} t + C_2\]Using the initial condition \( \theta(0) = 0 \), solving gives:\[0 = \frac{1}{90} \times 1 - \frac{1}{12} \times 0 + C_2\]\[0 = \frac{1}{90} + C_2\] Thus, \[C_2 = -\frac{1}{90}\] So, \[\theta = \frac{1}{90} (8t + 1)^{5/2} - \frac{1}{12} t - \frac{1}{90}\]

The final expression for the angular displacement \( \theta \) in terms of time \( t \) is given by:\[\theta = \frac{1}{90} (8t + 1)^{5/2} - \frac{1}{12} t - \frac{1}{90}\] This expression accounts for the initial conditions and the integration of the given acceleration function.