How long will you chase?

This problem can be solved easily if the angular velocity of the two boys are equal about the center of the circle.

Now in polar coordinates square of total velocity ( $V$ ) = Sum of the squares of radial ( $V_{\phi}$ )+Tangential velocity (( $V_{t}$ ) of the body.

So $V^{2}=V_{\phi}^2+V_{t} ^2$

Let at time 't' the boy(who is chasing) be at a distance 'x' from the center.

So $V_{t}=x\omega$

Now $\omega=\frac{u}{R}$ (Remember $\omega$ is same for both boys and $v=u$ )

So $\sqrt { v^{ 2 }-{ \left( \frac { xv }{ R } \right) }^{ 2 } } =V_{ \phi }$

or $\sqrt { v^{ 2 }-{ \left( \frac { xv }{ R } \right) }^{ 2 } } =\frac { dx }{ dt }$

or $\frac { v }{ R } t=\int _{ 0 }^{ 28 }{ \frac { dx }{ \sqrt { { R }^{ 2 }-{ x }^{ 2 } } } }$

So $\frac { v }{ R } t=\frac { \pi }{ 2 }$

On solving we get $t=11sec$

I actually did this a little differently.

First, notice that "The second boy always moves towards the first boy" necessarily implies that, if $\vec{v_1}$ is the velocity of the boy at the center and $\vec{v_2}$ is the velocity of the boy at the edge, we must have:

$\vec{v_1} \cdot \vec{v_2} =0$ since the two are always normal. If we let:

$\vec{v_2} = (v\cos(\frac{vt}{r}), v\sin(\frac{vt}{r}))$ and let:

$\vec{v_1} = (\frac{dx}{dt}, \frac{dy}{dt})$

It becomes clear that: $\frac{dx}{dt} = -\tan(\frac{vt}{r}) \frac{dy}{dt}$

If we use the additional facts that $\parallel \vec{v_1} \parallel = v$ and $x(0) = 0$ , $y(0)=0$ , we eventually arrive at:

$\frac{vt}{r} = \frac{\pi}{2}$ which immediately leads to $t=11$

Also, it can be seen that at any instant the second boy's direction is towards the first boy. Hence, by symmetry the path described by the second boy will also be a circle of R= 14m. and he will catch the first boy by travelling half of this circle. Hence T= ((22/7) * 14 )/4 = 11sec. Note: the tangent drawn to the new circle will always direct towards second boy.

The problem can be made a little more interesting if the boy on the circular path can change his direction at any time. The end result is nonetheless the same.