Answer:
4.83m/[tex]s^{2}[/tex]
Explanation:
For a particle moving in a circular path the resultant acceleration at any point is the vector sum of radial and the tangential acceleration
Radial acceleration is given by [tex]a_{radial}=w^{2}[/tex]r
Applying values we get [tex]a_{radial}=(2t)^{2}[/tex]X0.3m
Thus [tex]a_{radial}=1.2t^{2}[/tex]
At time = 2seconds [tex]a_{radial}= 4.8m/s^{2}[/tex]
The tangential acceleration is given by [tex]a_{tangential} =\frac{dV}{dt}=\frac{d(wr)}{dt}[/tex]
[tex]a_{tangential}=\frac{d(2tr)}{dt}[/tex]
[tex]a_{tangential}= 2r[/tex]
[tex]a_{tangential}=0.6m/s^{2}[/tex]
Thus the resultant acceleration is given by
[tex]a_{res} =\sqrt{a_{rad}^{2}+a_{tangential}^{2}}[/tex]
[tex]a_{res} =\sqrt{4.8^{2}+0.6^{2} } =4.83m/s^{2}[/tex]
