Question

A particle moves in a circle of radius 30 cm. It's linear speed varies with time as shown in figure. What will be it's acceleration at t = 3s ?

Answer 1

By graph⬇️
$\tan( \frac{\pi}{3} ) = \frac{v}{t} \\ \frac{ \sqrt{3} }{2} = \frac{v}{t} \\ v = \frac{ \sqrt{3}t }{2} \\ on \: diffrentiating \: both \: sides \\ \frac{dv}{dt} = \frac{ \sqrt{3} }{2} \\ a = \frac{ \sqrt{3} }{2} \\ \\ angular \: acceleration \alpha = \frac{a}{r} \\ \alpha = \frac{ \sqrt{3} }{2 \times r} \\ \alpha = \frac{ \sqrt{3} }{60} cm {s}^{ - 2}$
Hope it will help you✔️✔️

Answer 2

∆ Non - Uniform Circular Motion ∆

Heya Mate

¥ For a Non - Uniform Circular Motion, you must know,

$\vec{a} = ( \frac{ - {v}^{2} }{r} ) \hat{r} + (\frac{dv}{dt} ) \hat{r_t}$

where,

( i ) v is a function of "t"

( ii ) And the components are along and perpendicular to the radius vector of the trajectory of motion

Now, ^^"
$\tan( \frac{\pi}{3} ) = \sqrt{3}$

( is indeed a Tricky Step :p )

So, we have :

$v(t) = \sqrt{3} t$

And so,

$\vec{a} = ( - \frac{ {( \sqrt{3}t) }^{2} }{ 30 } ) \hat{r} + ( \sqrt{3} ) \hat{r_t}$

At t = 3s, we have :

$\vec{a} = ( - \frac{9}{10} ) \hat{r} + ( \sqrt{3} ) \hat{r_t}$

And so,

$| \vec{a}| = \sqrt{3.81} \: cm {s}^{ - 2}$

Apologies if the Answer goes wrong but from the next time ^^" be sure to Plot the Graph with Units Specified ( cm/s or m/s for speed )