Answer
$$
a_{t}=6.0 \mathrm{m} / \mathrm{s}^{2}
$$
Work Step by Step
The angular acceleration is, from Eq. $10-8$ ,
$$
\alpha=\frac{d \omega}{d t}=\frac{d}{d t}(0.60 t)=0.60 \mathrm{rad} / \mathrm{s}^{2} \text { . }
$$
Then, the tangential acceleration at $t=5.0 \mathrm{s}$ is, using $\mathrm{Eq}$ . $10-22$ ,
$$
a_{t}=r \alpha=(10 \mathrm{m})\left(0.60 \mathrm{rad} / \mathrm{s}^{2} \right)=6.0 \mathrm{m} / \mathrm{s}^{2}
$$
