Math, asked by gmahara58, 6 months ago

At t1 =2.00s the acceleration of a particle is counter clockwise circular motion is (5√3) ICAP+(5)jcap.It moves at constant speed. At time t2= 5.00s,the particles acceleration is (5)icap+(-5√3)jcap.What is the radius of the path taken by the particle if t2-t1 is less than one period?

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Answered by yadavsushant282

Step-by-step explanation:

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At t1 =2.00s the acceleration of a particle is counter clockwise circular motion is (5√3) ICAP+(5)jcap.It moves at constant speed. At time t2= 5.00s,the particles acceleration is (5)icap+(-5√3)jcap.What is the radius of the path taken by the particle if t2-t1 is less than one period?​

Answers

At t1 =2.00s the acceleration of a particle is counter clockwise circular motion is (5√3) ICAP+(5)jcap.It moves at constant speed. At time t2= 5.00s,the particles acceleration is (5)icap+(-5√3)jcap.What is the radius of the path taken by the particle if t2-t1 is less than one period?