Question

A particle moves in a circle of radius 5 cm with constant speed and time period 0.2πs. The acceleration of the particle is :​

Answer 1

Hope this helps you..

Answer 2

Answer:

The acceleration of the particle is $5\ m/s^2$

Explanation:

Given that,

Radius of circle $r= 5\ cm$

Time period $T = 0.2\pi\ sec$

We know that,

The  velocity is defined as,

$v = \dfrac{2\pi r}{T}$

Where, v = velocity

r = radius

T = time period

Put the value into the formula

$v = \dfrac{2\times3.14\times5\times10^{-2}}{0.2\times3.14}$

$v=0.5\ m/s$

Now,

The acceleration is defined as

$a =\dfrac{v^2}{r}$

Where, a = acceleration

v = velocity

Put the value into the formula

$a = \dfrac{0.5^2}{5\times10^{-2}}$

$a = 5\ m/s^2$

Hence, The acceleration of the particle is $5\ m/s^2$