a person moves from town A to town B, 4km away in the direction N30E of A. then the person moves due east of B to a town C, 5km away and finally due south to town D, 3km away from c.

Question

(a) find the distance covered
(b) how would you find the displacement

Answer 1

To solve the problem, let's break down the movements step by step and visualize each segment.

(a) Finding the Distance Covered

1. From Town A to Town B:

   • The distance from A to B is 4 km.

2. From Town B to Town C:

   • The movement from B to C is due east, covering a distance of 5 km.

3. From Town C to Town D:

   • The movement from C to D is due south, covering a distance of 3 km.

To calculate the total distance covered, we simply add the distances of each leg of the journey:

\[ \text{Total Distance} = \text{AB} + \text{BC} + \text{CD} \] \[ \text{Total Distance} = 4 , \text{km} + 5 , \text{km} + 3 , \text{km} = 12 , \text{km} \]

Answer for (a):

The total distance covered is 12 km.

(b) Finding the Displacement

Displacement is the straight-line distance from the initial point (Town A) to the final point (Town D), along with its direction. We will compute this vectorially by determining the coordinates of each town.

1. Town A: Let's place Town A at the origin, \( A(0, 0) \).

2. Town B:

   • Moving in the direction N30E means \( 30^\circ \) east of north.
   • We can calculate the coordinates of B:
     • \( x_B = 4 \sin(30^\circ) = 4 \times 0.5 = 2 \) km (eastward)
     • \( y_B = 4 \cos(30^\circ) = 4 \times \frac{\sqrt{3}}{2} = 4\sqrt{3}/2 \approx 3.46 \) km (northward)
   • Therefore, \( B(2, , 3.46) \).

3. Town C:

   • Moving due east from B 5 km, we get:
   • \( C(2 + 5, , 3.46) = C(7, , 3.46) \).

4. Town D:

   • Moving due south 3 km from C:
   • Coordinates for D will be \( D(7, , 3.46 - 3) = D(7, , 0.46) \).

Calculating Displacement:

Now we find the displacement vector \( \vec{AD} \): \[ \vec{AD} = D - A = (7 - 0, , 0.46 - 0) = (7, , 0.46) \]

The distance (magnitude of the displacement) can be calculated using the Pythagorean theorem:

\[ \text{Displacement} = \sqrt{(7)^2 + (0.46)^2} \] \[ \text{Displacement} = \sqrt{49 + 0.2116} \approx \sqrt{49.2116} \approx 7.01 , \text{km} \]

Direction of Displacement:

To find the angle (θ) of the displacement from the eastward direction: \[ \tan(\theta) = \frac{\text{Vertical}}{\text{Horizontal}} = \frac{0.46}{7} \] Calculating θ: \[ \theta = \tan^{-1}\left(\frac{0.46}{7}\right) \approx \tan^{-1}(0.0657) \approx 3.77^\circ \]

Answer for (b):

The displacement from Town A to Town D is approximately 7.01 km at an angle of 3.77° north of east.