Question 16 of 121

A person moved from $A$ to $B$ on a circular path as shown in figure. If the distance travelled by him is $60m$, then the magnitude of displacement would be : (Given $\cos 135^{\circ}=-0.7$)

A$42m$B$47m$C$19m$D$40m$

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Answer: (B) $47m$

The distance travelled by the person $d$ will be equal to the arch length.
$\Rightarrow d=R\theta$
$\Rightarrow 60=R\left(\dfrac{3\pi }{4}\right)$
$\Rightarrow R=\dfrac{60\times 4}{3\pi }=\dfrac{80}{\pi }m$
Now the magnitude of displacement can be found using subtraction of vectors. Therefore,
$\left|s\right|=\sqrt{{{R}_{2}}^{2}+{{R}_{1}}^{2}-2{R}_{1}{R}_{2}\cos \theta } \Rightarrow \left|s\right|=\sqrt{{R}^{2}+{R}^{2}-2{R}^{2}\cos 135^{\circ}}$
$\Rightarrow \sqrt{2{R}^{2}-2{R}^{2}\left(-0.7\right)}$
$\Rightarrow \sqrt{3.4{R}^{2}}=\sqrt{3.4{\left(\dfrac{80}{\pi }\right)}^{2}}$
$\Rightarrow R=47m$