# Question in the image below ?

Jul 16, 2018

$\Delta \boldsymbol{v} = {\boldsymbol{v}}_{B} - {\boldsymbol{v}}_{A}$

Using the scalar product:

$\left\mid \Delta \boldsymbol{v} \right\mid = \sqrt{\left({\boldsymbol{v}}_{B} - {\boldsymbol{v}}_{A}\right) \cdot \left({\boldsymbol{v}}_{B} - {\boldsymbol{v}}_{A}\right)}$

$= \sqrt{{v}_{B}^{2} + {v}_{A}^{2} - 2 \left\mid {\boldsymbol{v}}_{B} \right\mid \left\mid {\boldsymbol{v}}_{A} \right\mid \cos \theta}$

$= \sqrt{{V}^{2} + {V}^{2} - 2 {V}^{2} \cos \theta} q \quad \because \left\mid {\boldsymbol{v}}_{A} \right\mid = \left\mid {\boldsymbol{v}}_{B} \right\mid = V$

$\therefore \left\mid \Delta \boldsymbol{v} \right\mid = \sqrt{2} V \sqrt{1 - \cos \theta}$

For time $\Delta t$, distance travelled, $D$, is:

• $D = 2 \pi R \cdot \frac{{180}^{o} + {60}^{o}}{{360}^{o}} = \frac{4}{3} \pi R q \quad \because A \hat{O} R = {180}^{o} + {60}^{o}$

$\therefore \Delta t = \frac{D}{V} = \frac{4 \pi R}{3 V}$

Using the formula you are given:

$\left\mid {\boldsymbol{a}}_{\text{ave}} \right\mid = \frac{\left\mid \Delta \boldsymbol{v} \right\mid}{\Delta t}$

$= \frac{3 \sqrt{2} {V}^{2} \sqrt{1 - \cos \theta}}{4 \pi R}$

In the scalar product, $\theta$ is the angle between ${\boldsymbol{v}}_{A}$ and ${\boldsymbol{v}}_{B}$. From the geometry: $\theta = {120}^{o} , \cos \theta = - \frac{1}{2}$

$\left\mid {\boldsymbol{a}}_{\text{ave}} \right\mid = \frac{3 \sqrt{3} {V}^{2}}{4 \pi R}$

Dimensional analysis is always a useful check:

$\frac{{\left(L {T}^{- 1}\right)}^{2}}{L} = L {T}^{- 2}$ <-- Correct unit of acceleration