Question

A particle starts from rest and moves in a circle of radius 3m.Angular acceleration of the particle varies as Alpha=4t rad/sec^2.Then:
(A) particle is moving in a uniform circular motion
(B) particle will travel a distance of 2 m after 1 sec
(C) angle between the velocity and the acceleration vector will be 45 after 1 sec (D) after 1 sec speed of the particle will be 6 m/sec

Answer 1

Answer:

option d is correct after 1 sec speed of the particle will be 6 m/sec

Answer 2

After 1 sec speed of the particle will be 6 m/sec

(D) is correct option.

Explanation:

Given that,

Angular acceleration $\alpha=4t\ rad/s^2$

We need to calculate the angular velocity

Using formula of angular velocity

$\alpha=\dfrac{d\omega}{dt}$

Put the value into the formula

$4t=\dfrac{d\omega}{dt}$

$d\omega=4t dt$

On differentiating

$\omega=\dfrac{4t^2}{2}$

$\omega=2t^2$

We need to calculate the velocity

Using formula of velocity

$v=r\omega$

Put the value into the formula

$v=3\times2t^2$

When t = 1 sec,

$v=6\times1^2$

$v=6\ m/s$

Hence, After 1 sec speed of the particle will be 6 m/sec

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