Engineering Mechanics: Statics & Dynamics (14th Edition)

$a_n=35.6m/s^2$ $a_t=2.83m/s^2$
We can determine the required normal and tangential components of acceleration as follows: $\alpha=5t^{\frac{1}{2}}$ $\implies \alpha=5()^{\frac{1}{2}}=7.07rad/s^2$ Similarly $d\omega=\alpha dt$ $\implies \int_0^{\omega} d\omega=\int_0^t 5t^{\frac{1}{2}}dt$ This simplifies to: $\omega=\frac{5}{3/2}t^{\frac{3}{2}}$ At $t=2s$ $\implies \omega=\frac{10}{3}(2)^{1.5}$ $\implies \omega=9.42rad/s$ Now the normal component of the acceleration is given as $a_n=\omega^2 r$ We plug in the known values to obtain: $a_n=(9.42)^2\times 0.4$ $\implies a_n=35.6m/s^2$ and the tangential component of the acceleration is $a_t=\alpha r$ We plug in the known values to obtain: $a_t=(7.04)(0.4)$ $\implies a_t=2.83m/s^2$