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  • In the given figure represents the total acceleration of a particle moving in the clockwise direction

In the given figure a = 15 \frac{m}{s^2}represents the total acceleration of a particle moving in the clockwise direction in a circle of radius R = 2.5{\rm m} at a given instant of time. The speed of the particle is

Option: 1

4.5{{\rm m} \mathord{\left/ {\vphantom {{\rm m} {\rm s}}} \right. \kern-\nulldelimiterspace} {\rm s}}


Option: 2

5.0{{\rm m} \mathord{\left/ {\vphantom {{\rm m} {\rm s}}} \right. \kern-\nulldelimiterspace} {\rm s}}


Option: 3

5.7{{\rm m} \mathord{\left/ {\vphantom {{\rm m} {\rm s}}} \right. \kern-\nulldelimiterspace} {\rm s}}


Option: 4

6.2{{\rm m} \mathord{\left/ {\vphantom {{\rm m} {\rm s}}} \right. \kern-\nulldelimiterspace} {\rm s}}


Answers (1)

best_answer

Given :    R=2.5 m                   a=15  m/s2

From figure, centripetal acceleration      

a_c = a cos30^o = 15 \times 0.866 = 13   m/s2

Using       a_c=\frac{v^2}{R}

∴   13 =\dfrac{v^2}{2.5}                  

   ?v=5.7   m/s

Posted by

seema garhwal

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