Deriving an equation for acceleration in circular motion

Question

I have a question:

A particle starts to move from rest in a circle of radius 3m, so after $t$ seconds its speed is $5t+1$m/s. Find its acceleration after 1 second.

I have tried differentiating $5t+1$ to give $5$ but this doesn't make and sense as a constant acceleration would always give an increasing radius to the circle.

I have also looked into polar coordinate acceleration ie $$\ddot r_v= (\ddot r - r \dot \theta^2) \hat r_v + (2 \dot r\dot \theta + r \ddot \theta) \hat \theta_v$$ And tried to work out the angular acceleration $\hat \theta_v$ with the information given in the question, but that didn't make any sense either.

Answer 1

Hint:

I suppose that $5t+1$ is the modulus of velocity.

Use: $ x=R\cos \theta\quad ,\quad y=R\sin \theta$ so that: $$ \dot x=-R\dot \theta \sin \theta \qquad \dot y=R\dot \theta \cos \theta $$ find: $$ |v|=R\dot \theta=5t+1 $$ So you have $\dot \theta$ and you can find $\theta, \ddot \theta$ ....

Answer 2

Well, we know that for a circular path, we have two types of acceleration, radial, and tangential. The radial acceleration can be computed as $a_r = \frac{v_t^2}{r} = \frac{(5t + 1)^2}{3} = \frac{25t^2 + 10t + 1}{3}$. And the tangential acceleration would be simply 5 as you computed it. So to find the resulting total acceleration, we just use the Pythagorean theorem, since the components are perpendicular. So $ a = \sqrt{ (\frac{25t^2 + 10t + 1}{3})^2 + 5^2} $ and since we want an acceleration at $t=1$s, we can say $a=\sqrt{(\frac{36}{3})^2+25}=\sqrt{144+25}=\sqrt{169}=13$. So your acceleration would actually be 13 m/s^2.