Question

a particle is moving on a circular path of radius 1 m with 2m/s. If speed starts increasing at a rate of 2m/s^2 then the acceleration of particle is?

Answer 1

The acceleration has two components tangential acceleration and centripetal acceleration.
Tangential acceleration at=2m/s2
Centripetal acceleration ar=v2r=(2)21=4m/s2
Total acceleration is a=(2)2+(4)2−−−−−−−−−√=4.47m/s2

Regards.

Answer 2

Answer:

The acceleration of particle is 4.47 m/s².

Explanation:

Given that,

Radius = 1 m

velocity = 2 m/s

If speed starts increasing at a rate of 2m/s^2.

The acceleration has two components tangential acceleration and centripetal acceleration.

.The tangential acceleration is given

$a_{t}= 2\ m/s^2$

The centripetal acceleration is given by,

$a_{c}=\dfrac{v^2}{r}$

$a_{c}=\dfrac{2^2}{1}$

$a_{c}=4 m/s^2$

The acceleration of the particle is

$a=\sqrt{(a_{t})^2+(a_{c})^2}$

$a=\sqrt{(2)^2+(4)^2}$

$a = 4.47\ m/s^2$

Hence, The acceleration of particle is 4.47 m/s².