# Starting from rest, a particle is constrained to move in a circle of radius 4 m. The tangential acceleration is a_t = 9 m/s^2. How long will it take to rotate 45º?

Mar 1, 2017

$t = \sqrt{\frac{2 \pi}{9}} \text{seconds}$

#### Explanation:

If you think of this as a linear problem, the magnitude of the velocity will simply be:
$| v | = | {v}_{0} | + | a \cdot t |$
And the other equations of motion work in a similar way:
$d = {v}_{0} \cdot t + \frac{1}{2} a \cdot {t}^{2}$

The distance along the direction of travel is simply one eighth of a circle:
$d = 2 \pi \cdot \frac{r}{8} = 2 \pi \cdot \frac{4}{8} = \pi \text{ meters}$

Replacing this value in the equation of motion for distance gives:
$\pi = {v}_{0} \cdot t + \frac{1}{2} a \cdot {t}^{2}$
$\pi = 0 \cdot t + \frac{1}{2} a \cdot {t}^{2}$
$2 \pi = a \cdot {t}^{2}$
$2 \pi = 9 \cdot {t}^{2}$
$\frac{2 \pi}{9} = {t}^{2}$
$\sqrt{\frac{2 \pi}{9}} = t$