Question

A particle is moving on a circular path of radius R=15/pi m with constant speed 10m/s . If it moves from a to b as shown in figure then magnitude of average acceleration of the particle during its journey will be

Answer 1

Answer:

2.12 m/s^2

Explanation:

Assuming a and b as same point thus the displacement is zero.

radius of the circular path R= 15/pi m

speed = 10 m/s

Centripetal acceleration a_c= $\frac{v^2}{r}$

= $\frac{10^2}{15/pi}$

=2.12 m^2/s

Tangential acceleration (a_t) is equal to

= Change in velocity/ time take            ( velocity is constant)

= 0/T =     0

so the tangential acceleration is zero.

therefore, average acceleration is only due to centripetal one.

thus  magnitude of average acceleration of the particle during its journey will be  = 2.12 m/s^2

Answer 2

Answer:

2.12 m/s^2.......

HOPE IT HELPS....!!!