Answer
$a_n=35.6m/s^2$
$a_t=2.83m/s^2$
Work Step by Step
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:
$a_t=(7.04)(0.4)$
$\implies a_t=2.83m/s^2$
