Question

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

• A. 42 m
• B. 47 m
• C. 19 m
• D. 40 m

Answer 1

The Correct Option is B

To find the magnitude of displacement, we need to determine the straight-line distance between the initial point \( A \) and the final point \( B \) on the circular path. The displacement in this circular motion is the chord length of the arc, \( AB \).

Given:

• Angle subtended at the center \( \angle AOB = 135^\circ \).
• Arc length \( AB = 60 \text{ m} \).

We can use the formula for the arc length:

s = r\theta

where \( s \) is the arc length, \( r \) is the radius, and \( \theta \) is the angle in radians.

First, convert \( 135^\circ \) to radians:

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

Now, solve for the radius \( r \):

60 = r \times \frac{3\pi}{4}

r = \frac{60 \times 4}{3\pi}

r \approx \frac{240}{3 \times 3.14} \approx 25.48 \text{ m}

The displacement is the chord length \( AB \), which can be calculated using the cosine rule in the triangle \( OAB \):

AB^2 = OA^2 + OB^2 - 2 \times OA \times OB \times \cos(135^\circ)

AB^2 = 2r^2 (1 - \cos(135^\circ))

Since \( \cos(135^\circ) = -0.7 \), substitute the values:

AB^2 = 2 \times (25.48)^2 \times (1 + 0.7)

AB^2 = 2 \times 648.23 \times 1.7 \approx 2204

AB \approx \sqrt{2204} \approx 46.95 \text{ m}

The closest option is \( 47 \text{ m} \).

Therefore, the magnitude of the displacement is 47 m.