Question

A particle moves in a circular path of radius R such
that its speed v varies with distance x as v = avx ,
where a is a positive constant. The acceleration of
the particle after traversing a distance V2 R is
​

Answer 1

Answer:

Explanation:

Correct statement:

A particle moves in a circular path such that its speed v varies with distance  s  as  v = √ s , where  α  is  α  positive constant. Find the acceleration of the particle after traversing a distance s .

Total acceleration

a  = √ a^ 2 t + a^ 2 r  

= √ ( d v d t ) ^2 + ( v ^2 R ) ^2

where  

v = α √ s

Differentiating  v = α √ s with respect to time, we have

d v/ d t =  α  s − 1 / 2/ 2 d s d t

Substituting  

d s /d t  =  α √ s , we have  d v /d t = α 2 /2

Now substituting  d v / d t  and  v  in the expression of  a , we have  

a =  √ ( α ^2 /2 ) ^2  +  [ α √ s /R ] ^2

This gives  

a =α 2 √ 1 /4 + s ^2 R ^2

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