Question

74. A particle moves in a circular path of radius R
with an angular velocity o=a-bt where a and
b are positive constants and t is time. The
magnitude of the acceleration of the particle after
time 2a/bis​

Answer 1

Explanation:

given Ш = a-bt

We know that

•  v =r Ш = R (a-bt )

The motion of the particle is circular

a = $\sqrt{ac +at}$

• ac = v² /R  = R (a -bt )²

Given t = 2a/b

ac = R ( a - 2a )² =  -Ra

• at = R α

• α = dШ /d t = -b

at = -Rb

a = R$\sqrt{a2 +b2}$

                           Hope you understand

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