Answer
Ans:
$$
\begin{aligned}
& v_E=3 \mathrm{~m} / \mathrm{s} \\
& \left(a_E\right)_t=2.70 \mathrm{~m} / \mathrm{s}^2 \\
& \left(a_E\right)_n=600 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$
Work Step by Step
$$
\begin{aligned}
& \omega_A r_A=\omega_B r_B \\
& \omega_A(10)=\omega_B(40) \\
& \omega_B=\frac{1}{4} \omega_A \\
& v_E=\omega_B r_E=\frac{1}{4} \omega_A(0.015)=\left.\frac{1}{4}(40)\left(t^3+6 t\right)(0.015)\right|_{t=2} \\
& v_E=3 \mathrm{~m} / \mathrm{s} \\
& \alpha_A=\frac{d \omega_A}{d t}=\frac{d}{d t}\left[40\left(t^3+6 t\right)\right]=120 t^2+240 \\
& \alpha_A r_A=\alpha_B r_B \\
& \alpha_A(10)=\alpha_B(40) \\
& \alpha_B=\frac{1}{4} \alpha_A \\
& \left(a_E\right)_t=\alpha_B r_E=\left.\frac{1}{4}\left(120 t^2+240\right)(0.015)\right|_{t=2} \\
& \left(a_E\right)_t=2.70 \mathrm{~m} / \mathrm{s}^2 \\
& \left(a_E\right)_n=\omega_B^2 r_E=\left.\left[\frac{1}{4}(40)\left(t^3+6 t\right)\right]^2(0.015)\right|_{t=2} \\
& \left(a_E\right)_n=600 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$
