# PROBLEM 10 A point describes a circumference according to law s (t) = t³ + 2t² where s measured meters along the circumference and, in seconds. If a total acceleration of the point is 16 2 m / s2, when t = 2 s.calculate the radius R of the circumference?

Apr 15, 2018

$R = 25 \text{ m}$

#### Explanation:

The acceleration vector (radial and tangential) is:

$m a t h b f a = - \frac{{\dot{s}}^{2}}{R} m a t h b f \setminus m a t h b f \hat{r} + \ddot{s} \setminus m a t h b f \hat{\theta}$

Computing derivatives:

$s \left(t\right) = {t}^{3} + 2 {t}^{2}$

$\dot{s} \left(2\right) = 3 {t}^{2} + 4 t {|}_{t = 2} = 20$

$\ddot{s} \left(2\right) = 6 t + 4 {|}_{t = 2} = 16$

$m a t h b f a \left(2\right) = - \frac{400}{r} m a t h b f \setminus m a t h b f \hat{r} + 16 \setminus m a t h b f \hat{\theta}$

CORRECTING TYPO IN QUESTION:

$| m a t h b f a \left(2\right) | = 16 \textcolor{red}{\sqrt{2}} = \sqrt{{\left(- \frac{400}{R}\right)}^{2} + {\left(16\right)}^{2}}$

Solves as: $R = 25 \text{ m}$