Circular Motion #4

The tangential acceleration of a particle moving in a circular path is $\frac{ dv } {dt }$ , that is the derivative of speed of the particle with respect to time. In the problem,

$a_{T} = \frac{ dv } { dt} = \frac{ 4 \pi } { 6 } \cos \left( \frac{\pi } { 6 } t\right)$

When $t = 5$ , $a_{T} = \dfrac{ 4 \pi } { 6 } \cos \left( \dfrac{5 \pi } { 6 } \right) = \dfrac{ 4 \pi } { 6 } \times \dfrac{ - \sqrt {3} } { 2}$ .

The magnitude of tangential acceleration is $| a_{T} | = \dfrac{ 4 \pi } { 6 } \times \dfrac{\sqrt {3} } { 2} \approx \boxed{ 1.814 }$ $_\square$