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Question

The tangential acceleration of a particle in a circular motion of radius 2 m is at=αt m/s2 (where α is a constant) . Initially, the particle is at rest. Net acceleration vector of the particle makes 45 with the radial acceleration after 2 sec. The value of constant α is:

A
12 m/s3
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B
1 m/s3
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C
2 m/s2
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D
1 m/s2
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Solution

The correct option is B 1 m/s3
Tangential acceleration is the rate of change of speed along circular path.
at=dvdt ...(i)
Given:-
at=αt m/s2
dvdt=αt
Integrating both sides:
v=vv=0dv=αt=2t=0tdt
v=α[t22]20=2α
Speed at t=2 s is v=2α

At t=2 s, the acceleration vectors can be represented as:


From vector triangle at t=2s :
tan45=atac
at=ac
(αt)=v2R (2α)=(2α)22
α=1 m/s3

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