Question

One afternoon, a couple walks three-fourths of the way around a circular lake, the radius of which is 2.08 km. They start at the west side of the lake and head due south to begin with. (a) What is the distance they travel? (b) What is the magnitude of the couple’s displacement? (c) What is the direction (relative to due east) of the couple’s displacement?

Answer 1

Answer:

Explanation:

a ) Distance travelled by them = 3 / 4 x 2π R where R is radius of the circular path .

= .75 x 2 x 3.14 x 2.08

= 9.8 km approx

b )

Displacement = shortest distance between final and initial position

= √2 R

= 1.414 x 2.08

= 2.94 km approx

c ) The direction of displacement is towards north east

so the direction is 45° north of east .

Answer 2

Answer:

(a). The distance is 9.80 km.

(b). The magnitude of the couple’s displacement is 2.94 km.

(c). The direction of the couple’s displacement is 45°.

Explanation:

Given that,

Radius of lake = 2.08 km

(a). We need to calculate the total distance

Using formula of distance

$d=\dfrac{3}{4}(2\pi R)$

Put the value into the formula

$d=\dfrac{3}{4}(2\pi\times2.08)$

$d=9.80\ km$

(b). We need to calculate the magnitude of the couple’s displacement

Using formula of displacement

$D=\sqrt{R^2+R^2}$

$D=\sqrt{2R^2}$

$D=\sqrt{2\times(2.08)^2}$

$D=2.94\ km$

(c), The direction of the displacement is given by

Using formula of direction

$\tan\theta=\dfrac{R}{R}$

$\theta=\tan^{-1}(1)$

$\theta=45^{\circ}$

Hence, (a). The distance is 9.80 km.

(b). The magnitude of the couple’s displacement is 2.94 km.

(c). The direction of the couple’s displacement is 45° relative to the east.