A particle is moving in a circle of radius R in such a way that at any instant the normal and the tangential component of its acceleration are equal. If its speed at t=0 is v₀ then the time it takes to complete the first revolution is R v₀ (1-e^ - ) . Find the value of ( + )^6 .

Step-by-step solution

aligned & d v d t = v^2 R _ v₀ ^v d v v^2 = 1 R ₀^t d t (- 1 v )_ v₀ ^v= t R v= v₀ 1- v₀ R t d s d t = v₀ 1- v₀ R t ₀^ 2 R d s= ₀^t v₀ (1- v₀ R t ) d t & 2 R=-R [ (1- v₀ R t ) ]₀^t 2 =- (1- v₀ R t ) 1- v₀ R t=e^ -2 t= R v₀ (1-e^ -2 ) & a=1, =2 ( + )^6=(1+2)^6=729 aligned