# What will be the anser?

Aug 11, 2018

$5.7 \frac{m}{s}$

#### Explanation:

Here,considering speed to be the magnitude of linear velocity.

So,if it is moving with constant speed of $v$ at a certain point of time,then its Centripetal acceleration is $| {\vec{a}}_{c} | = {v}^{2} / r$ where $r$ is the radius of the circular path.

If its linear acceleration is ${\vec{a}}_{r}$ then,

${\vec{a}}_{c} + {\vec{a}}_{r} = \vec{a}$

Given, $| \vec{a} | = 15 m {s}^{-} 2$

Again angle between ${\vec{a}}_{c}$ and ${\vec{a}}_{r}$ is ${90}^{\circ}$

So, ${a}^{2} = {a}_{r}^{2} + {a}_{c}^{2}$

Or, ${a}^{2} = {a}_{r}^{2} + {v}^{4} / {r}^{2.} . . 1$

Given, $a = 15 m {s}^{-} 2 , r = 2.5 m$

If, $\vec{a}$ makes an angle of ${30}^{\circ}$ w.r.t ${\vec{a}}_{c}$ then,

$\tan 30 = | {\vec{a}}_{r} \frac{|}{|} {\vec{a}}_{c} |$

I.e $\frac{1}{\sqrt{3}} = {a}_{r} / \left({v}^{2} / r\right)$

Given, $r = 2.5 m$

So, putting the values and arranging we get,

${a}_{r} = {v}^{2} / \left(2.5 \sqrt{3}\right)$

Putting this value of ${a}_{r}$ in the previous equation $1$ we get,

$v = 5.7 \frac{m}{s}$

Aug 11, 2018

Acceleration vector in non-uniform circular motion is:

• $\boldsymbol{a} \left(t\right) = {\underbrace{r \dot{\theta} \setminus {\boldsymbol{\hat{e}}}_{\theta}}}_{\text{tangential ") - underbrace(v^2/r\ bb hat e_r)_("inward radial: } = {\boldsymbol{a}}_{r}}$

In given geometry:

$\left\mid {\boldsymbol{a}}_{r} \right\mid = \left\mid \boldsymbol{a} \right\mid \cos 30$

$\implies {v}^{2} / 2.5 = 15 \frac{\sqrt{3}}{2}$

$\therefore \text{speed" = abs(bbv) ~~ 5.7 " m/s}$