Question

A particle moves in a circle of radius 3m with a velocity v = 4t^2i + 3t^2j what is its centripetal acceleration at the end of 2 sec

Answer 1

A particle moves in a circle of radius 3m with a velocity v = 4t^2i + 3t^2j what is its centripetal acceleration at the end of 2 sec.

answer : 20m/s²

explanation : A particle moves in a circle of radius 3m with a velocity , v = 4t² i + 3t² j

we have to find centripetal acceleration at the end of 2sec.

acceleration is the rate of change of velocity with respect to time.

so, centripetal acceleration, $a_r$ = dv/dt

= d(4t² i + 3t² j)/dt

= {d(4t²)/dt} i + {d(3t²)/dt} j

= 4 × 2t i + 3 × 2t j

= 8t i + 6t j

at the end of t = 2sec, $a_r$ = (8 × 2i + 6 × 2 j) = 16i + 12 j

hence, magnitude of centripetal acceleration, $|a_r|$ = √(16² + 12²) = 20 m/s²

Answer 2

Answer:

133.2m/s^2

Explanation:

centipetal acceleration= v^2/radius

v= 4*4i+ 3*4j

= 16i+12j

|v|= √16^2+ 12^2

= 20 m/s

Ac= acceleration centripetal = 20*20/3=133.2