Question

Please solve q 12?

Answer 1

Let at certain instant the revolving particle be at position `A` and after half revolution of the circular path it reaches diametrically opposite point at `B`.If the oppositely directed velocities of the particle at these two poitions be represented by `V_A andV_B` respectively then let `V_A=-5`m/s
and `V_B=+5`m/s

So change in velocity `DeltaV=+5-(-5)=10` m/s

The path length from A to B = half of the perimeter `=pi*5`m.

The time required to traverse this path with speed 5m/s will be `Deltat=(pi*5)/5=pi` s.

Hence average acceleration

`a=(DeltaV)/(Deltat)=10/pi` `ms^-2`