A person moved from A to B on a circular path as shown in figure. If the distance travelled by him is 60…

The distance travelled by the person $d$ will be equal to the arch length.

$\Rightarrow d=R\theta$

$\Rightarrow 60=R\left(\frac{3\pi }{4}\right)$

$\Rightarrow R=\frac{60\times 4}{3\pi }=\frac{80}{\pi }\mathrm{m}$

Now the magnitude of displacement can be found using subtraction of vectors. Therefore,

$$\begin{gathered} \left|s\right|=\sqrt{{R_2}^2+{R_1}^2-2R_1{R}_2\cos \theta } \\ \Rightarrow \left|s\right|=\sqrt{R^2+R^2-2R^2\cos 135°} \end{gathered}$$

$\Rightarrow \sqrt{2R^2-2R^2\left(-0.7\right)}$

$\Rightarrow \sqrt{3.4R^2}=\sqrt{3.4{\left(\frac{80}{\pi }\right)}^2}$

$\Rightarrow R=47 \mathrm{m}$