Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Solve it, A particle is moving in a circular path of radius R with velocity v =kt, where k is a constant, then acceleration of the particle at t = t is

A particle is moving in a circular path of radius R with velocity v = kt, where k is a constant, then acceleration of the particle at t = t is 

  • Option 1)

    a = \sqrt{\frac{k^4 t^4}{R^2}+ \frac{k^2}{4}}

  • Option 2)

    a = \sqrt{\frac{k^2 t^2}{2R^2}+ {k^2}}

  • Option 3)

    a = \sqrt{\frac{k^4 t^4}{R^2}+ {k^2}}

  • Option 4)

    a = \sqrt{\frac{k^4 t^4}{R^2}+ {4k^2}}

 

Answers (1)

best_answer

As we learnt,

 

Circular Motion -

a=\sqrt{{a_{t}}^{2}+\left (\frac{v^{2}}{r} \right )^{2}}
 

- wherein

a_{t}= tangential acceleration

\frac{v^{2}}{r}= centripetal acceleration

 

 \\*a_t = \frac{\mathrm{d} v}{\mathrm{d} x} = k \\*r_r = \frac{v^2}{R} = \frac{k^2t^2}{R} \\*a = \sqrt{a^{2}_{t} + a^{2}_{r}} = \sqrt{K^2 +( \frac{k^2t^2}{R})^2} = \sqrt{k^2 + \frac{k^4t^4}{R^2}}

 


Option 1)

a = \sqrt{\frac{k^4 t^4}{R^2}+ \frac{k^2}{4}}

Option 2)

a = \sqrt{\frac{k^2 t^2}{2R^2}+ {k^2}}

Option 3)

a = \sqrt{\frac{k^4 t^4}{R^2}+ {k^2}}

Option 4)

a = \sqrt{\frac{k^4 t^4}{R^2}+ {4k^2}}

Posted by

Avinash

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question