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

We need to find the magnitude of the displacement of a person who moves from point $$A$$ to point $$B$$ along a circular path, covering a distance of $$60\text{ m}$$.

1. Relate Distance to the Radius ($$R$$) of the Circular Path

From the problem , let's assume the angle subtended by the arc $$AB$$ at the center of the circular path is $$\theta = 135^\circ$$.

The arc length (distance traveled) is given by the formula:

$$\text{Distance} = R\theta \quad (\text{where } \theta \text{ is in radians})$$

Convert the angle from degrees to radians:

$$\theta = 135^\circ \times \frac{\pi}{180^\circ} = \frac{3\pi}{4}\text{ radians}$$

Substitute the given distance ($$60\text{ m}$$) to find the radius ($$R$$):

$$60 = R \times \frac{3\pi}{4}$$

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

2. Calculate the Magnitude of Displacement

The straight-line distance (displacement, $$d$$) between two points on a circle of radius $$R$$ separated by an angle $$\theta$$ can be calculated using the law of cosines or the standard shortcut formula:

$$d = 2R \sin\left(\frac{\theta}{2}\right)$$

Alternatively, using the given value of $$\cos 135^\circ = -0.7$$ directly via the vector subtraction formula:

$$d = \sqrt{R^2 + R^2 - 2R^2 \cos\theta} = \sqrt{2R^2(1 - \cos 135^\circ)}$$

Substitute $$\cos 135^\circ = -0.7$$ into the equation:

$$d = \sqrt{2R^2(1 - (-0.7))} = \sqrt{2R^2(1 + 0.7)} = \sqrt{2R^2(1.7)} = \sqrt{3.4R^2}$$

$$d = R\sqrt{3.4}$$

3. Substitute $$R$$ to Find the Final Value

Now, substitute $$R = \frac{80}{\pi}$$ and approximate $$\pi \approx 3.14$$ along with $$\sqrt{3.4} \approx 1.844$$:

$$d = \left(\frac{80}{3.14}\right) \times 1.844$$

$$d \approx 25.48 \times 1.844 \approx 47\text{ m}$$

Conclusion

The magnitude of the displacement is approximately 47 m, which corresponds to Option B.