Return of the Particle Race

First I started with Q, according to given condition

$R\frac { d\omega }{ dt } =\frac { { v }^{ 2 } }{ R }$ (Let the velocity of particle at any arbitrary point be $v$ )

So

$\frac { d\omega }{ dt } ={ \omega }^{ 2 }$

On integrating we get

$-\frac { 1 }{ \omega } =t+C$

At $t=0$ $\omega=\frac{v_{0}}{R}$

So $C=-\frac{R}{v_{0}}$

So the equation becomes

$-\frac { 1 }{ \omega } =t-\frac{R}{v_{0}}$

So on substituting $\omega=\frac{d\theta}{dt}$ we get

$\int _{ 0 }^{ T }{ \frac { dt }{ \frac { R }{ { v }_{ 0 } } -t } } =\int _{ 0 }^{ \pi }{ d\theta }$

So, $\ln { \left( \frac { R }{ v } -t \right) } =-\pi$

or $\left( \frac { R }{ v } -t \right) { =e }^{ -\pi }$

On putting values I got $t=0.756$

Now for Particle P

It follows a projectile motion with range = $2R$

$\frac { { v }^{ 2 }sin(2\theta ) }{ g } =2R$ .........(1)

And Time of flight of projectile

$T_{0}=\frac { 2vsin(\theta ) }{ g }$

On putting value and using eq(1)

I got $T_{0}=1.76$

So it is clear that particle Q won the race so the answer is $1.76-0.75+5.31=6.32$ .