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

Option: 1
$42\: \mathrm{m}$

Option: 2
$47\: \mathrm{m}$

Option: 3
$19\: \mathrm{m}$

Option: 4
$40\: \mathrm{m}$

Answers (1)

best_answer

Distance

$\mathrm{ACB}=\mathrm{r\left ( 135^{o}\times \frac{\pi }{180} \right ) }$

$60=\frac{3}{4}\: \mathrm{\pi r}$

$\mathrm{r=\frac{80}{\pi }}\rightarrow (1)$

Magnitude of displacement $\mathrm{AB}$ is $\mathrm{S_{AB}=\sqrt{r^{2}+r^{2}-2r^{2}\cos 135^{0}}}$

(from cosine laws)

$\mathrm{S_{AB}=\sqrt{2r}\times \sqrt{1.7}}$

$\mathrm{S_{AB}=\sqrt{3.4}\: \: r}$

From eq.1

$\mathrm{S_{AB}=\sqrt{3.4}\times \frac{80}{\pi }}$

$=47\: \mathrm{m}$

Hence (2) is correct option.

Posted by

HARSH KANKARIA

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